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THOMSON'S    NEW    GRADtD    SERIES. 


NEW 


PRACTICAL 


AEITHMETIC: 


FOR 


GRAMMAR    DEPARTMENTS. 


BY  JAMKS  B.  THOMSON,  LI,  D., 

AUTHOR  OF  DAT  A  THOMSON'S  ARITHMETICAL  8ERIES  ;  EDITOR  OP  DAT'S  SCHOOi 
ALGEBRA,  LEQENDRE'S  GEOMETRY,  ETC. 


THIRTY-FIFTH    EDITION. 


NEW     YORK: 

CLARK  &  MAYNARD,  PUBLISHERS, 

5  BARCLAY  STREET. 
CHICAGO:    46  MADISOX  STREET. 


TIJOIJSON'S  NEW  GRADED  SERIES. 

IN    THREE    BOOKS. 

I.     NEW    MENTAL    ARITHMETIC. 
(For  Primary  Departments.) 

I      NEW     RUDIMENTS     OF     ARITH- 
METIC.   (For  Intermediate  Departments.) 

III.    NEW   PRACTICAL  ARITHMETIC 
(For  Grammar  Departments.) 

KEY    TO    NEW    PRACTICAL   ARITH- 
METIC.    (For  Teachers  only.) 

THOMSON'S    SUPPLEMENTARY    COURSE. 
FOR  HIGHER  INSTITUTIONS.    (In  preparation.) 


Entered  according  to  Act  of  Congress,  In  the  year  1872,  bv 

JAMES   B.    THOMSON, 
In  the  Office  of  the  Librarian  of  Congress,  at  Washington,  D.  C. 


Electrotyped  by  SMVTH  &  McDouoAi.  Ba  Beekman  Street,  N.  Y. 


03 


1877 
PREFACE. 


r  I  ^HE  New  Practical  Arithmetic  now  offered  to  the 
_L  public,  is  the  third  and  last  of  the  works  which 
constitute  the  author's  "  New  Graded  Series." 

The  old  "Practical"  was  issued  in  1845,  and  revised  in 
1853.  Since  that  time  important  changes  have  taken 
place  in  the  commercial  world.  These  changes  necessarily 
affect  business  calculations,  and  demand  corresponding 
modifications  in  text  books. 

To  meet  this  demand,  the  "New  Graded  Series"  was 
undertaken.  Each  part  of  the  series  has  been  reinvesti- 
gated  and  rewritten; — the  whole  being  readjusted  upon 
the  graded  plan,  and  brought  down  to  the  present  wants. 

Among  the  objects  aimed  at  in  the  present  work  are 
the  following : 

1.  To  make  the  definitions  clear,  concise,  and  com- 
prehensive. 

2.  To  present  the  principles  of  the  science  in  a  series 
of  distinct  and  consecutive  propositions. 

3.  To  lead  the  mind  of  the  pupil,  through  the  analysis 
of  the  examples  immediately   following  the   respective 
propositions,  to  discover  the  principles  by  winch  all  similar 
examples  are  solved,  and  enable  him  to  sum  up  the  prin- 
ciples thus  developed, into  a  brief,  comprehensive  rule. 

4.  The  "how"  and  the  "why"  are  fully  explained. 

5.  Great  pains  have  been  taken  to  ascertain  the  Standard 
Weights  and  Measures  authorized  by  the  Government;  and 
to  discard  from  the  Tables  such   denominations  as   are 
obsolete,  or  not  used  in  this  country.* 

*  Laws  of  Congress ;  Profs.  Hassler,  Bnche  and  Egleston ;  Reports  of  Supts. 
Harrison  and  Calkins ;  also  of  the  Committee  on  Weights  and  Meat-urea  of  the 
Fa:  is  Exposition,  1867. 

409537 


PllEP  A  CE. 

6.  The  Metric  System  is  accompanied  with  brief  and 
appropriate  explanations  for  reducing  it  to  practice.     Its 
simplicity  and  comprehensiveness  have  secured  its  use  in 
the  natural  sciences  and  commerce  to  such  an  extent  in 
this  and  foreign  countries,  that  no  student  can  be  said  to 
have  a  finished  education,  without  a  knowledge  of  it. 

7.  Particular  attention  has  been  paid  to  the  develop- 
ment of  Analysis,  the  grand  common  sense  rule  which 
business    men    intuitively    adopt    as    they   enter    upon 
practical  life. 

8.  The  examples  are  new  and  abundant ; — being  drawn 
from  the  various  industrial  arts,  commerce,  science,  etc. 

9.  The  arrangement  of  the  matter  upon  the  page,  and  the 
typography,  have  also  received  due  attention.     Teachers 
who  deal  much  with  figures,  will  be  pleased  with  the 
adoption  of  the  Franklin  type.    The  ease  with  which 
these  figures  are  read,  is  sufficiently  attested  by  their  use 
in  all  recent  Mathematical  Tables. 

Finally,  it  has  been  the  cardinal  object  to  adapt  the 
science  of  numbers  to  the  present  wants  of  the  farm,  the 
household,  the  workshop,  and  the  counting-room ; — in  a 
word,  to  incorporate  as  much  information  pertaining  to 
business  forms,  and  matters  of  science,  as  the  limits  of  the 
book  would  permit.  In  this  respect  it  is  believed  the 
work  is  unrivaled.  While  it  puts  forth  no  claim  to 
mathematical  paradoxes,  it  is  believed  teachers  will  find 
that  something  worthy  of  their  attention  is  gained,  in 
nearly  every  Article. 

In  conclusion,  the  author  tenders  his  most  cordial 
thanks  to  teachers  and  the  public  for  the  very  liberal 
patronage  bestowed  upon  his  former  Arithmetics,  known 
as  "Day  and  Thomson's  Series."  It  is  hoped  the  "Ne\v 
Graded  Series"  will  be  found  worthy  of  continued  favor. 

JAMES  B.  THOMSON. 
NEW  YORK,  July.,  1872 


CONTENTS. 


PAS* 

Number,       ....--..9 

Notation,  -                       -  10 

Arabic  Notation,  -                                                  -  10 

Roman  Notation,       -  15 

To  Express  Numbers  by  Letters,   -                         -  1 7 

Numeration,  -  17 

French  Numeration,     -                                            -  18 

English  Numeration,  20 

Addition,     -  -21 

When  the  Sum  of  each  Column  is  less  than  10,  -  23 

When  the  Sum  of  a  Column  is  10  or  more,    -        -  24 

Carrying  Illustrated,  -  24 

Drill  Columns,      -                                                  -  29 

Subtraction,    -  31 
When  each  Figure  in  the  Subtrahend  is  less  than 

that  above  it,-                                                     -  32 
When   a  Figure  in  the   Subtrahend  is  greater 

than  that  above  it,       -  34 

Borrowing  Illustrated,  -                                            -  34 

Questions  for  Review,  38 

Multiplication,                                                -  40 

When  the  Multiplier  has  but  one  Figure,   -  43 

When  the  Multiplier  has  more  than  one  Figure,    -  45 

To  find  the  Excess  of  93,    -  47 

Contractions,                                   -        -        -        -  49 


VI  CONTENTS. 

FAGS 

Division,  -  -  -  53 

The  Two  Problems  of  Division,  -  55 

Short  Division,  -  '57 

Long  Division,  -  6 1 

Contractions,  -  65 

Questions  for  Review,  69 

General  Principles  of  Division,  -  -  7 1 
Problems  and  Formulas  in  the  Fundamental  Rules,  7  2 

Analysis,  77 

Classification  and  Properties  of  Numbers,  -  -  80 

The  Complement  of  Numbers,  -  82 

Divisibility  of  Numbers,  83 

Factoring,  -  85 

Prime  Factors,  -  -  86 

Cancellation,  -  88 

Greatest  Common  Divisor,  -  -  91 

Least  Common  Multiple,  -  96 

Fractions,  -  -    99 

To  find  a  Fractional  Part  of  a  Number,  -        -       102 

General  Principles  of  Fractions,     -  -  103 

Reduction  of  Fractions,      -  -       104 

A  Common  Denominator,      -  -  1 1 1 

The  Least  Common  Denominator,     -  -       112 

Addition  of  Fractions,  -  -  114 

Subtraction  of  Fractions,   -  117 

Multiplication  of  Fractions,  -  -120 

General  Rule  for  multiplying  Fractions,  -        -       125 

Division  of  Fractions,    -  -126 

General  Rule  for  dividing  Fractions,  -       i ;,  i 

Questions  for  Review,    -                          -  -  132 

Fractional  Relations  of  Numbers,       -  134 

Decimal  Fractions,  -  139 

Reduction  of  Decimals,      ...  -       144 

Addition  of  Decimals,   -        -        -        -  -        -146 

Subtraction  of  Decimals,  -  -  ...  148 
Multiplication  of  Decimals,  -  ...  -  149 

Division  of  Decimals,                  -        -  -        -       152 


CONTENTS.  Vii 

PAGE 

United  States  Money,     -  154 

Addition  of  U.  S.  Money,  -        -  -                158 

Subtraction  of  U.  S.  Money,  -        -        -  -  159 

Multiplication  of  U.  S.  Money,   -        -  -                160 

Division  of  U.  S.  Money,        -        -        -  -  1 6 1 

Counting-room  Exercises,  -     •    -        -  -                163 

Making  out  Bills,  -        -        -        -        -  -         -164 

Business  Methods,     -  166 

Compound  Numbers, 171 

Money, 171 

Weights, 175 

Measures  of  Extension,      ...  .       177 

Measures  of  Capacity,    -  -  182 

Circular  Measure,       -        -        -        -  -        -184 

Measurement  of  Time,  -        -  -  186 

Reduction,                           -  189 

Application  of  "Weights  and  Measures,  -  -     194 

Artificers'  Work,       ...  -         196 

Measurement  of  Lumber,     -  -     196 

Denominate  Fractions,     -  202 

Metric  Weights  and  Measures,  -  -  207 

Application  of  Metric  Weights  and  Measures,  -       2 1  .\ 

Compound  Addition,     -  -216 

Compound  Subtraction,     -        -  -219 

Compound  Multiplication,     -  -  224 

Compound  Division,  -  -       226 

Comparison  of  Time  and  Longitude,      -  -  227 

Percentage^     -  -      230 

Notation  of  Per  Cent,    -                          -        -  -  230 

Five  Problems  of  Percentage,     -                 -  233 

Application*  of  Percentage,  -  241 

Commission  and  Brokerage,       -        -        -  -      241 

Account  of  Sales, 24^ 

Profit  and  Loss, 247 

Interest,  

Preliminary  Principles,     -  255 

Six  Per  Cent  Method,  -        -  256 


CONTENTS. 

PAOl 


Method  by  Aliquot  Parts,      -  ...  259 

Method  by  Days,        ......      26o 

Partial  Payments,  -  267 

Compound  Interest,  -  -       273 

Discount,       -  -        ...  .  276 

Banks  and  Bank  Discount,  -       278 

•Stock  Investments,      -  -  280 

Government  Bonds,          -  ...        28i 

Exchange,  -  -    285 

Insurance,          -        -        -        -        -  -292 

Taxes,  -------  -  295 

Duties,      --....-.       298 

Internal  Revenue,          ...  .  300 

Equation  of  Payments,       -        -        -  301 

Averaging  Accounts,     ...  -  304 

Ratio,  ....  .      307 

Proportion,        ....  .          309 

Simple  Proportion,    -        -        -  -        -311 

Simple  Proportion  by  Analysis,     -  -313 

Compound  Proportion,       -  -      316 

Partitive  Proportion,     ...  -  319 

Partnership,       -        -        ...  -      320 

Bankruptcy,-        -        ....  .  323 

Alligation,          -        -        -        -        -        -        -324 

Tnvolution,  -  ......  330 

Formation  of  Squares,  -  -332 

Evolution,  -  .....  333 

Extraction  of  the  Square  Root,  ...      335 

Applications  of  Square  Root,  -  339 

Formation  of  Cubes,  ....  .      343 

Extraction  of  the  Cube  Root,         -  345 

Applications  of  Cube  Root,        -        ...      349 

Arithmetical  JProaression,   -       -       -          350 

(Geometrical  Progression,  -      353 

Mensuration,    .......  355 

Miscellaneous  Examples,    -  ...      360 


ARITHMETIC. 


Art.  1.   Arithmetic  is  the  science  of  numbers. 

Arithmetic  is  sometimes  said  to  be  both  a  science  and  an  art :  a 
science  when  it  treats  of  the  theory  and  properties  of  numbers ;  an 
art  when  it  treats  of  their  applications. 

NOTES. — i.  The  term  arithmetic,  is  from  the  Greek  arithmStike, 
the  art  of  reckoning. 

2.  The  term  science,  from  the  Latin  scientia,  literally  signifies 
knowledge.  In  a  more  restricted  sense,  it  denotes  an  orderly  ar- 
rangement of  the  facts  and  principles  of  a  particular  branch  of 
knowledge. 

2.  Nnnibcf  is  a  imit,or  a  collection  of  units. 

A  Unit  is  any  single  thing,  called  one.  One  and 
one  more  are  called  tu-o ;  two  and  one  more  are  called 
three;  three  and  one  more,  four,  etc.  The  terms  one, 
two,  three,  four,  are  properly  the  names  of  numbers,  but 
are  often  used  for  numbers  themselves. 

NOTE. — The  term  unit  is  from  the  Latin  unus,  signifying  one. 

3.  The   Unit  Otie  is  the  standard  by  which  all  num- 
bers are  measured.    It  may  also  be  considered  the  base 
or  element  of  number.     .For,  all  whole  numbers  greater 
than  one  are  composed  of  ones.    Thus,  two  is  composed  of 
one  and  one.     Three  is  one  more  than  two;  but  two,  we 
have  seen,  is  composed  of  ones;  hence,  three  is,  and  so  on. 


QUESTIONS. — i.  What  is  arithmetic?     What  else  is  it  sometimes  said  to  be? 
When  a  science  ?    An  art  ?    2.  Number  ?    A  unit  ?    3.  The  Lomdard 
numbers  are  measured  ?    The  base  or  element  of  number  ? 


10  NOTATION. 

4.   Numbers  are  either  abstract  or  concrete. 

An  Abstract  Number  is  one  that  is  not  applied  to 
any  object;  as,  three,  five,  ten. 

A  Concrete  Number  is  one  that  is  applied  to  some 
object;  as,  five  peaches,  ten  books. 


NOTATION. 

5.  Notation   is  the  art  of  expressing  numbers  by 
figures,  letters,  or  other  numeral  characters. 

The  two  principal  methods  in  use  are  the  Arabic  and 
the  Roman. 

NOTE. — Numbers  are  also  expressed  by  words  or  common  lan- 
guage ;  but  this,  strictly  speaking,  is  not  Notation. 

ARABIC   NOTATION. 

6.  The  Arabic  Notation  is  the  method  of  express- 
ing numbers  by  certain  characters  called  figures. 

It  is  so  called,  because  it  was  introduced  into  Europe 
from  Arabia. 

7.  The  Arabic  figures  are  the  following  ten,  viz  : 

1,      2,      3,      4,      5,      6,      7,      8,      9,      O. 

one,     two,    three,   four,    five,       six,  seven,  eight,    nine,  naught. 

The  first  nine  are  called  significant  figures,  or  digits ; 
the  last  one,  naught,  zero,  or  cipher. 

NOTES. — i.  The  first  nine  are  called  significant  figures,  because 
each  always  expresses  a  number. 

2.  The  term  digit  is  from  the  Latin  digitus,  a  finger,  and  was 
applied  to  these  characters  because  they  were  employed  as  a  substi 


4.  What  is  an  abstract  number?  Concrete?  5.  What  is  notation?  The 
principal  methods  in  use?  6.  Arabic  notation  ?  Why  so  called  ?  7.  TIow  many 
figures  does  it  employ  ?  What  are  the  first  nine  called  ?  The  last  one  ?  Note. 
Why  called  significant  figures ?  Why  digits?  Meaning  of  digitu?  ?  Why  is  th» 
Jtt*t  called  naught  ?  Meaning  of  zero  ?  Of  cipher  ? 


NOTATION.  11 

tute  for  the  fingers  upon  which  the  ancients  used  to  reckon.  The 
term  originally  included  the  cipher,  but  is  now  generally  restricted 
to  the  first  nine. 

3.  The  last  one  is  called  naught;  because,  when  standing  alone,  it 
has  no  value,  and  when  connected  with  significant  figures,  it  denotes 
the  absence  of  the  order  in  whose  place  it  stands. 

4.  Zero  is  an  Italian  word,  signifying  nothing. 

The  term  cipher  is  from  the  Arabic  sifr  or  sifreen,  empty,  vacant. 
Subsequently  the  term  was  applied  to  all  the- Arabic  figures  indis, 
crimiuately ;  hence,  calculations  by  them  were  called  ciphering. 

8.  Each  of  the  first  nine  numbers  is  expressed  by  a 
single  figure, — each  figure  denoting  the  number  indicated 
by  its  name.  These  numbers  are  called  units  of  the  first 
order,  or  simply  units. 

8,  a.  ^Nine  is  the  greatest  number  expressed  by  one  figure. 
Numbers  larger  than  nine  are  expressed  thus: 

Ten  (i  more  than  9)  is  expressed  by  an  ingenious  de- 
vice, which  groups  ten  single  things  or  ones  together,  and 
considers  the  collection  a  new  or  second  order  of  units, 
called  ten.  Hence,  ten  is  expressed  by  writing  the  figure  i 
in  the  second  place  with  a  cipher  on  the  right;  as,  10. 

The  numbers  from  ten  to  nineteen  inclusive  are  ex- 
pressed by  writing  i  in  the  second  place,  and  the  figure 
denoting  the  units  in  the  first ;  as, 
u,      12,       13,       14,       15,      16,        17,         18,        19. 

eleven,  twelve,  thirteen,  fourteen,  fifteen,  sixteen,  seventeen,  eighteen,  nineteen. 

Twenty  (2  tens)  is  expressed  by  writing  the  figure  2  in 
the  second  place  with  a  cipher  on  the  right;  as,  20. 

Thirty  (3  tens)  by  writing  3  in  the  second  place  with  a 
cipher  on  the  right ;  and  so  on  to  ninety  inclusive ;  as, 

20,        30,       40,        50,       60,         70,        80,        90. 

twenty,      thirty,       forty,          fifty,         sixty,      seventy,     eighty,      ninety. 


8.  How  are  the  firat  nine  numbers  expressed  ?  What  are  they  called  ?  What 
is  the  greatest  number  expressed  by  one  figure  ?  How  is  ten  expressed  ?  Twenty  ? 
Thirty,  etc.  ?  The  numbers  between  10  and  so?  From  20  to  99  inclusive  ? 


12  NOTATION. 

The  numbers  from  twenty  to  thirty,  and  so  on  to  ninety^ 
nine  (99)  inclusive,  are  expressed  by  writing  the  tens  in  the 
second  place,  and  the  units  in  the  first ;  as, 

21,        22,        23,        34,       35,        46,        57,       99. 

twenty-     twenty-     twenty-      thirty-      thirty-        forty-        fifty-       ninety- 
one,  two,         three,        four,        five,          six,          seven,       nine. 

8,  b.  Ninety -nine  is  the  greatest  numbed  that  can  be 
expressed  by  two  figures. 

A  hundred  (i  more  than  99)  is  expressed  by  grouping 
ten  units  of  the  second  order  together,  and  forming  a  new 
or  third  order  of  units,  called  a  hundred.  Thus,  a  hun- 
dred is  expressed  by  writing  i  in  the  third  place  with  two 
ciphers  on  the  right;  as,  100. 

In  like  manner,  the  numbers  from  one  hundred  to  nine 
hundred  and  ninety-nine  inclusive,  are  expressed  by  writ- 
ing the  hundreds  in  the  third  place,  the  tens  in  the  second, 
and  the  units  in  the  first.  Thus,  one  hundred  and  thirty- 
five  (i  hundred,  3  tens,  and  5  units,)  is  expressed  by  135. 

8,  c.  Nine  hundred  and  ninety-nine  is  the 

greatest  number  that  can  be  expressed  by  three  figures. 

Tliousands,  and  larger  numbers,  are  expressed  by  form- 
ing other  new  orders,  called  the  fourth,  fifth,  etc.,  orders ; 
as,  tens  of  thousands,  hundreds  of  thousands,  millions,  etc. 

NOTES. — i.  The  names  of  the  first  ten  number?,  one,  two,  three, 
etc.,  are  primitive  words.  The  terms  eleven  and  twelve  are  from  the 
Saxon  endlffen  and  ticelif,  meaning  one  and  ten,  two  and  ten.  Thir- 
teen is  from  thir  and  teen,  which  mean  three  and  ten,  and  so  on. 

2.  Twenty  is  from  the  Saxon  tweentig,  tween,  two,  and  ty,  tens ; 
i.  e.,  two  tens.     Thirty  is  from  thir  and  ty,  three  tens,  and  so  on. 

3.  The  terms  hundred,  thousand,  and  million  are  primitive  words, 
having  no  perceptible  analogy  to  the  numbers  they  express. 

From  the  foregoing  illustrations  we  derive  the  following  principle : 
8.  How  is  a  hundred  expressed  T    Thousands,  and  larger  numbers  t 


NOTATION.  It 

9.  The  Orders  of  Units  increase  by  the  scale  of  ten 
That  is,  ten  single  units  are  one  ten  ;  ten  tens  one  hun- 
dred; ten  hundreds  one  thousand  ;  and,  universally, 

Ten  of  any  lower  order  make  a  unit  of  the  next 
higher. 

NOTES. — i.  If  the  term  unit  denotes  one,  how,  it  may  be  asked,  can 
ten  things  or  ones  be  a  unit,  ten  tens  another  unit,  etc.  And  how- 
can  the  figures  i,  2,  3,  etc.,  sometimes  denote  single  things  or  ones  ; 
at  others,  tens  of  ones,  and  so  on. 

The  answer  is,  units  are  of  two  kinds,  simple  and  collective. 

A  simple  unit  is  a  single  thing  or  one. 

A  collective  unit  denotes  a  group  of  ones,  regarded  as  a  whole. 

2.  To  illustrate  these  units,  suppose  a  basket  of  pebbles  is  before 
us.  Counting  them  out  one  by  one,  each  pebble  is  a  simple  unit. 

Again,  counting  out  ten  single  pebbles  and  putting  them  togethei 
in  a  group,  this  group  forms  a  unit  of  the  second  order,  called  ten. 
Counting  out  ten  such  groups,  and  putting  them  together  in  one 
pile,  this  collection  forms  a  unit  of  the  third  order,  called  hundred. 
In  like  manner,  a  group  of  ten  hundreds  forms  a  unit  of  the  fourth 
order,  called  thousand,  and  so  on. 

These  different  groups  of  ten  are  called  units  on  the  same  principle 
that  a  group  of  ten  cents  forms  a  unit,  called  a  dime  ;  or,  a  group  of 
ten  dimes,  a  unit,  called  a  dollar.  Thus,  it  will  be  seen  that  the 
figures  i,  2,  3,  4,  etc.,  always  mean  one,  two,  three,  four  units 
as  their  name  indicates ;  but  the  value  of  these  units  depends  upon 
the  place  the  figure  occupies. 

10.  From  the  preceding  illustrations,  it  will  be  seen 
that  the  Arabic  Notation  is  founded  upon  the  following 
principles : 

ist.  Numbers  are  divided  into  groups  called  units,  of 
the  first,  second,  third,  etc.,  orders. 

zd.  To  express  these  different  orders  of  units,  a  simple 
and  a  local  value  are  assigned  to  the  significant  figures, 
according  to  the  place  they  occupy. 

3d.  If  any  order  is  wanting,  its  place  is  supplied  by  a 
cipher. 

9.  How  do  the  orders  of  units  increase?  Units  make  a  ten  ?  Tens  a  hundred' 
to.  Name  the  principles  upon  which  the  Arabic  Notation  in  founded  ? 


14  NOTATION. 

11.  The  Simple  Value  of  a  figure  is  the  number 
of  units  it  expresses  when  it  stands  alone,  or  in  the  right- 
hand  place. 

The  Local  Value  is  the  number  it  expresses  when 
connected  with  other  figures,  and  is  determined  by  the 
place  it  occupies,  counting  from  the  right. 

12.  It  is  a  general  law  of  the  Arabic  Notation  that  the 
value  of  a  figure  is  increased  tenfold  for  every  place  it  is 
moved  from  the  right  to  the  left ;  and,  conversely, 

The  value  of  each  figure  is  diminished  tenfold  for  every 
place  it  is  moved  from  the  left  to  the  right.  Thus,  2  in 
the  first  place  denotes  two  simple  units ;  in  the  second 
place,  ten  times  two,  or  twenty ;  in  the  third  place,  ten 
times  as  much  as  in  the  second  place,  or  two  hundreds, 
and  so  on.  (Art.  8.) 

13.  The  number  denoting  the  scale  by  which  the  orders 
of  units  increase  is  called  the  radix. 

The  radix  of  the  Arabic  Notation  is  ten ;  hence  it  is 
often  called  the  decimal  notation. 

NOTES. — i.  The  term  radix,  Latin,  signifies  root,  or  base.  The 
term  decimal  is  from  the  Latin  decent,  ten. 

2.  The  decimal  radix  was  doubtless  suggested  by  the  number  of 
fingers  (digiti)  on  both  hands.  (Art.  7,  Note.)  Hence, 

14.  To  Express  Numbers  by  Figures, 

Begin  at  the  left  hand  and  write  the  figures  of  the  given 
orders  in  the  successive  places  toward  the  right. 

If  any  intermediate  orders  are  omitted,  supply  their 
places  ivith  ciphers. 


ii.  The  simple  value  of  a  figure?  Local?  12.  What  is  the  law  as  to  mov'ig 
a  figure  to  the  right  or  left  ?  13.  What  is  the  radix  of  a  system  of  notation  ?  Tho 
radix  of  the  Arabic  ?  What  else  is  the  Arabic  system  called  ?  Why  ?  Note. 
Meaning  of  radix  ?  Decimal?  What  so£"-p!>t«d  the  decimal  radix?  14.  Rule 
for  expressing  numbers  by  figures  ? 


NOTATION.  5 

EXERCISES. 

Express  Uie  following  numbers  by  figures: 

1.  Three  hundred  and  forty-five. 

2.  Four  hundred  and  sixty. 

3.  Eight  hundred  and  four. 

4.  Two  thousand  three  hundred  and  ten. 

5.  Thirty  thousand  and  nineteen. 

6.  Sixty-three  thousand  and  two  hundred. 

7.  One  hundred  and  ten  thousand  two  hundred  and 
twelve. 

8.  Four  hundred  and  sixty  thousand  nine  hundred 
and  thirty. 

9.  Six  hundred  and  five  thousand  eight  hundred  and 
forty-two. 

10.  Two  millions  sixty  thousand  and  seventy-five. 

ROMAN    NOTATION. 

15.  The  Roman  Notation  is  the  method  of  ex- 
pressing numbers  by  certain  letters.    It  is  so  called  because 
it  was  employed  by  the  Romans.      The  letters  used  are 
the  following  seven,  viz. :  I,  V,  X,  L,  C,  D  and  M.      The 
letter  I  denotes  one;   V,ftve;   X,  ten;   L,  fifty;  C,  one 
hundred  ;  D,  five  hundred ;  M,  one  thousand.    Interven- 
ing and  larger  numbers  are  expressed  by  the  repetition 
and  combination  of  these  letters. 

16.  The  Roman   system  is  based  upon  the  following 
general  principles : 

ist.  It  proceeds  according  to  the  scale  of  ten  as  far 
as  a  thousand,  the  unit  of  each  order  being  denoted  by 

15.  Roman  notation?  Why  so  called?  Letters  employed?  The  letter  I  de- 
note? V?  X?  L?  C?  D?  M?  :6.  Name  the  first  principle  upon  which  it 
Is  based.  What  is  the  effect  of  repeating  n  letter  ?  Of  placing  a  letter  of  less 
r:ilue  before  one  of  greater  value?  If  placed  after?  If  a  line  is  placed  over  a 
letter? 


16 


NOTATION. 


a  single  letter.  Thus,  I  denotes  one ;  X,  ten ;  C,  one  hun- 
dred ;  M,  one  thousand. 

2d.  Repeating  a  letter,  repeats  its  value.  Thus,  I  denotes 
one ;  II,  two ;  III,  three ;  X,  ten ;  XX,  twenty,  etc. 

3d.  Placing  a  letter  of  less  value  before  one  of  greater 
value,  diminishes  the  value  of  the  greater  by  that  of  the 
less ;  placing  the  less  after  the  greater,  increases  the  value 
of  the  greater  by  that  of  the  less.  Thus,  V  denotes  five, 
but  IV  denotes  only  four,  and  VI  six. 

4th.  Placing  a  horizontal  line  over  a  letter  increases  its 
value  a  thousand  times.  Thus,  I  denotes  a  thousand ;  X, 
ten  thousand ;  0,  a  hundred  thousand ;  M,  a  million. 


TABLE. 


I, 

denotes  one. 

XXX,    denotes  thirty. 

n, 

"       two. 

XL,             "       forty. 

in, 

"       three. 

L,               "       fifty. 

IV, 

"       four. 

LX,             "       sixty. 

V, 

"       five. 

LXX,          "       seventy. 

VI, 

"       six. 

LXXX,       "       eighty. 

VII, 

"       seven. 

XC,             "       ninety. 

VIII, 

"       eight. 

C,                "       one  hundred. 

IX, 

"       nine. 

CC,              "       two  hundred. 

x, 

"       ten. 

CCC,           "       three  hundred. 

XI, 

"       eleven. 

CCCC,         "       four  hundred. 

XII, 

"       twelve. 

D,                "       five  hundred. 

XIII, 

"       thirteen. 

DC,             "       six  hundred. 

XIV, 

"       fourteen. 

DCC,           "       seven  hundred. 

xv, 

"       fifteen. 

DCCC,        "       eight  hundred. 

XVI, 

"       sixteen. 

DCCCC,      "       nine  hundred. 

XVII, 

"       seventeen. 

M,               "       one  thousand. 

XVIII, 

"       eighteen. 

MM,            "       two  thousand. 

XIX, 

"       nineteen, 

MDCCCLXXI,  denotes  one  thous- 

XX, 

"       twenty. 

and  eight  hundred  and  seventy-one. 

NOTES. — i.  Four  was  formerly  denoted  by  IIII;  nine  by  VIIII; 
forty  by  XXXX ;  ninety  by  LXXXX ;  Jive  hundred  by  10 ;  and  a 
thousand  by  CL>. 


NOTATION. 


17 


2.  Annexing  0  to  1,0  (five  hundred)  increases  its  value  ten  times. 
Thus,  100  denotes  five  thousand ;  1000,  fifty  thousand. 

Prefixing  a  C  and  annexing  a  0  to  CIO  (a  thousand)  increases  its 
value  ten  times.  Thus,  CC10J  denotes  ten  thousand,  etc.  Hence, 

17.  To  Express  Numbers  by  Letters, 

Begin  at  the  left  hand  or  highest  order,  and  write  the 
letters  denoting  the  given  number  of  each  order  in  succession. 

NOTE. — The  Roman  Notation  is  seldom  used,  except  in  denoting 
chapters,  sections,  heads  of  discourses,  etc. 

EXERCISES. 
Express  the  following  numbers  by  letters  :  • 


i.  Fourteen, 
4.  Sixty-six, 
7.  Eighty-eight, 

10.   107, 

13-  613, 

1 6.  724, 

19.   1004, 

22.     1417, 
25.    1748, 


2.  Twenty-nine, 

5.  Forty-nine, 

8.  Ninety-four, 

II.  212, 

14-  S0?, 

17.  829, 

20.  1209, 

23.  1614, 

26.  1803, 


3.  Thirty-four, 
6.  Seventy-three, 
9.  Ninety-nine, 

12.  498, 

15.  608. 

18.  928, 

21.  1363, 

24.  1671, 

27.  1876. 


NUMERATION. 

18.  Numeration  is  the  art  of  reading  numbers  ex- 
pressed by  figures,  letters,  or  other  numeral  characters. 

NOTE. — The  learner  should  be  careful  not  to  confound  Numeration 
with  Notation.  The  distinction  between  them  is  the  same  as  that 
between  reading  and  vsritiny. 

19.  There  are  two  methods  oi'  reading  numbers,  the 
French  and  the  English. 


17.  Role  for  expressing  numbers  byletterc?  18.  Numeration?  Note.  Dis- 
tinction between  Numeration  and  Notation?  ig.  How  many  methods  of  readiujj 
numbers  V 


18  NUMERATION. 


FRENCH    NUMERATION. 

20.  The  French  divide  numbers  into  periods  of  three 
figures  each,  and  then  subdivide  each  period  into  units, 
tens,  and  hundreds,  as  in  the  following 

TABLE. 


r3  GO  S 

•5  §  £  .2  1  .§ 

«3—  ^3to  rf.  So5  2    P 

<§|  £j  s§     s§     ?i 

e^     ^<      t£  ti^    »— ' 


«tj  C?  __  r__ 

H  *§        «" 


O     §     ~  O   ^  O  = 

'S<2i>^     -S^oj     -S  w    .      iS "  ^ 


e 


S<"^  aJv-S  3t*H«  S'wS  V  ^  S  ® 

Sog  So=a  !-o~  ^og  ^os  C 

1§1  l§i  Is!  l§li  lal  1§1 

Sa»  !^oJ>  So^  3ol£  3opS  =Ot5 

WEHO>.  W  ft  w  WHH^  WEH^  KH^  WH^ 

823  561  729  452  789  384 

6th  period.  5th  period.  4th  period.  3d  period.  2d  period.  1st  period. 


The  first  period  on  the  right  is  called  units  period ;  the 
second,  thousands;  the  third,  millions,  etc. 

The  periods  in  the  table  are  thus  read  :  823  quadrillions, 
561  trillions,  729  billions,  452  millions,  789  thousand,  three 
hundred  and  eighty-four. 

NOTE. — The  terms  billion,  trillion,  quadrillion,  etc.,  are  derived 
from  the  Italian  milione  and  the  Latin  bis,  tres,  guatuor,  etc.  Thus, 
bis,  united  with  •million,  becomes  billion,  etc. 

21.  To  read  Numbers  according  to  the  French   Numeration. 

Divide  them  into  periods  of  three  figures  each,  counting 
from  the  right. 

•    Beginning  at  the  left  hand,  read  the  periods  in  succes- 
sion, and  add  the  name  to  each,  except  the  last. 


20.  The  French  method  ?  Repeat  the  Table,  bcgiuning  at  the  right.  What  is 
the  i Bt  period  called?  The  ad?  3d?  4th?  sth?  6th?  21.  The  rule  for  reading 
numbers  by  the  French  method  ?  Note  Why  omit  the  name  of  the  right  hand 
period  ?  The  difference  between  orders  and  periods  ? 


NUMERATION. 


19 


NOTES. — i.  The  name  of  the  right-hand  period  is  omitted  for  the 
sake  of  conciseness  ;  and  since  this  period  always  denotes  units,  the 
omission  occasions  no  obscurity. 

2.  The  learner  should  observe  the  difference  between  the  orders  of 
units  and  the  periods  into  which  they  are  divided.     The  former 
increase  by  tens ;  the  latter  by  thousands. 

3.  This  method  of  reading  numbers  is  commonly  ascribed  to  the 
French,  and  is  thence  called  FrencJi  numeration.    Others  ascribe  it  to 
the  Italians,  and  thence  call  it  the  Italian  method. 

Read  the  following  numbers  by  the  French  numeration  : 

21.  3006017 

22.  2460317239 

23.  5100024000 

24.  70300510 

25.  203019060 

26.  7800580019 

27.  86020005200 

28.  51036040 

29.  621000031 

30.  93000275320 


I. 

270 

ii. 

840230 

2. 

3°9 

12. 

4603400 

3- 

1270 

J3- 

35040026 

4- 

2036 

14. 

8o6oOOOO 

5- 

8605 

15- 

9001307 

6. 

40300 

1  6. 

65023009 

7- 

85017 

17- 

SlOOOO 

8. 

160401 

1  8. 

753O6O2O 

9- 

405869 

19. 

165380254 

10. 

1365406 

20. 

310400270 

31.  216327250516 

32.  4260300210109 

33.  300073004000 

34.  41295000400649 

35.  264300439000200 


36. 
37. 

38. 

39- 

40. 


Express  the  following  numbers  by  figures 

1.  Ten  millions,  five  thousand,  and  two  hundred. 

2.  Sixty-one  millions,  three  hundred  and  forty. 

3.  Three  hundred  and  ten  millions,  and  five  hundred. 

4.  Twenty-six  billions,  seventy  millions,  three  hundred. 

5.  One  hundred  billions,  four  hundred  and  twenty-five. 

6.  Sixty-eight  trillions,  seven  hundred  and  twenty-five. 

7.  Eight  hundred  and  twenty  millions,  five  hundred 
and  twenty- three. 


20  NUMERATION. 

8.  Sixty-seven  quadrillions,  ninety-seven  billions. 

9.  Four  hundred  and  sixty  quadrillions,  and  eighty, 
seven  millions. 

10.  Seven  hundred  and  sixty-one  quadrillions,  seventy- 
one  trillions,  two  hundred  billions,  eighteen  millions,  five 
thousand,  and  thirty-six. 

ENGLISH    NUMERATION. 

22.  The  English  divide  numbers  into  periods  of  six 
•figures  each,  and  then  subdivide  each  period  into  units, 
lens,  hundreds,  thousands,  tens  of .  thousands,  and  hundreds 
of  thousands,  as  in  the  following 

TABLE. 


<n  .rt 
II 


00 

a  o 

K    ^3 


^-33.2  «s|35    .  i| 

.Sc5f3a5  'ra^^d  -§    § 

°tj_C3tic»-;9  ®   *«    a     P  ^!     S  $<"§«> 

•   oS^OK  •o§t<o§  t,ogj^-         ^ 

s  M §  I  i  ijii^i  ^1^1 1 2 

407692  958604  413056 


3d  period.  id  period.  ist  period. 

The  periods  in  the  Table  are  thus  read :  407692  billions,  958604 
millions,  413  thousand,  and  fifty-six. 

NOTE. — This  method  is  called  English  numeration,  because  it  was 
invented  by  the  English. 

For  other  numbers  to  read  by  this  method,  the  pupil  is  referred 
to  those  in  Art.  21. 

as.  What  is  the  English  method  of  numeration  ?    Note.  Why  BO  called  T 


ADDITION. 

23.  Addition  is  uniting  two  or  more  numbers  in  one 
The  Sum  or  Amount  is  the  number  found  by  addi- 
tion.    Thus,  5   added  to  7  are  12;   twelve,  the  number 
obtained,  is  the  sum  or  amount. 

NOTES. — i.  The  sum  or  amount  contains  as  many  units  as  the 
numbers  added.  For,  the  numbers  added  are  composed  of  units ; 
and  the  whole  is  equal  to  the  sum  or  all  its  parts.  (Art.  3.) 

2.  When  the  numbers  added  are  the  same  denomination,  the 
operation  is  called  Simple  Addition. 

SIGNS. 

24.  Signs  are  characters  used  to  indicate  the  relation 
of  numbers,  and  operations  to  be  performed. 

25.  The   Sign    of  Addition  is   a    perpendicular 
cross  called  plus  (  +  ),  placed  before  the  number  to  be 
added.     Thus  7+5,  means  that  5  is  to  be  added  to  7,  and 
is  read  "  7  plus  5." 

NOTE. — The  term  plus,  Latin,  signifies  more,  or  added  to. 

26.  The  Sign  of  Equality  is  two  short  parallel 
lines  (  =  ),  placed  between  the  numbers  compared.     Thus 
7  +  5  =  12,  means  that  7  and  5  are  equal  to  12,  and  is  read, 
"  7  plus  5  equal  12,"  or  the  sum  of  7  plus  5  equals  12. 

Head  the  following  numbers : 


i.     8  +  4  +  2=6  +  8 


2-     7+3  +  5  = 

3.  19  +  1+0=6  +  5+   9 


4.  23+   7  =  19  +  11 

5.  37+   8=30  +  15 

6.  58+  10=40  +  28 


23.  What  is  addition  ?  The  result  called  ?  Note.  When  the  numbers  added 
are  the  same  denomination,  what  is  the  operation  called  ?  24.  What  are  Picons  ? 
25.  The  sifjn  of  addition  ?  Note.  The  meaning  of  plus?  26.  Sign  of  equality? 
How  is  7 +5 =12  read? 


22 


ADDITION. 


27.  The  Sign  of  Dollars  is  a  capital  S  with  two 
perpendicular  marks  across  it  ($),  prefixed  to  the  number 
of  dollars  to  be  expressed.  Thus,  $245  means  245  dollars. 

NOTE. — The  term  prefix,  from  the  Latin  prefigo,  signifies  to  place 
before. 


Read  the  following  expressions : 


1.  $17+ 

2.  $i3  + 

3.  $21+ 


=$i5+$io 


4- 
5- 
6.  $105  - 

ADDITION   TABLE. 


+  $8  = 


$36  -$96  +  $45 


i  and 

2  and 

3  and 

4  and 

5  and 

i  are  2 

i  are  3 

i  are  4 

i  are  5 

i  are  6 

2   «    3 

2   "    4 

2  «   5 

2   "    6 

2   «    7 

3  "   4 

3  "   5 

3  "   6 

3  "   7 

3  "   8 

4  "   5 

4  "   6 

4  "   7 

4  "   8 

4  «   9 

5  "  £ 

5  "   7 

5  "   8 

5  "   9 

5  "  10 

6  "   7 

6  «   8 

6  "   9 

6  "  10 

6  «  ii 

7  "   8 

7  "   9 

7  "  10 

7  "  ii 

7  «  12 

8  «   9 

8  "  10 

8  "  ii 

8   "   12 

8  «  13 

9  "  10 

9  "  ii 

9  "  12 

9  "  13 

9  «  14 

10   «  II 

10   "   12 

10   «   I3 

10   «  I4 

10  "  15 

6  and 

7  and 

8  and 

9  and 

10  and 

i  are  7 

i  are  8 

i  are  9 

i  are  10 

i  are  n 

2   "   8 

2  "   9 

2   "   10 

2   "   II 

2   "  12 

3  "   9 

3  "  10 

3  "  ii 

3  "  12 

3  "  13 

4  «  10 

4  "  ii 

4  "  12 

4  "  13 

4  "  14 

5  "  ii 

5  "  12 

5  "  13 

5  "  i4 

5  "  15 

6   "  12 

6  «  13 

6  «  14 

6  «  15 

6  "  16 

7  "  13 

7  "  14 

7  «  15 

7  "  16 

7  "  17 

8  "  14 

8  «  15 

8  "  16 

8  "  17 

8  «  18 

9  "  15 

9  "  16 

9  "  17 

9  "  18 

9  "  19 

10  "  16 

10  "  17 

10  "  18 

10  "  19 

10   ".  20 

More  mistakes  are  made  in  adding  than  in  any  other  arith- 
metical operation.  The  fiist  five  digits  are  easily  combined;  the 
results  of  adding  9,  being  i  less  than  if  10  were  added,  are  also  easy. 
The  othera  6,  7,  8,  are  more  difficult,  and  therefore  should  receive 
special  attention. 


-^    »7.  What  Is  the  Blgn  of  dollars  T 


ADDITION.  23 

CASE    I. 

28.  To  find  the  Amount  of  two  OP  more  numbers,  when 
the  Sum  of  each  column  is  Less  than  10. 

Ex.  i.  A  man  owns  3  farms;  one  contains  223  acres, 
another  51  acres,  and  the  other  312  acres:  how  many 
acres  has  he  ? 

ANALYSIS. — Let  the  numbers  be  set  down  as  in 

_.       .        .  OPERATION. 

the  margin.     Beginning  at  tlie  right,  we  proceed  tt- 

thus :  2  units  and  i  unit  are  3  units,  and  3  are  6  "H  g's 
units ;  the  sum  being  less  than  ten  units,  we  set  it 

under  the  column  of  units,  because  it  is  units.    Next,  2  23 

i  ten  and  5  tens  are  6  tens,  and  2  are  8  tens ;  the  5 1 

sum  being  less  than  10  tens,  we  set  it  under  the  ?  1 2 

column  of  tens,  because  it  is  tens.    Finally,  3  hun-  

dreds  and   2  hundreds  are   5  hundreds;   the  sum     A.ns.  586 
being  less  than  10  hundreds,  we  set  it  under  the 

column  of  hundreds,  for  the  same  reason.     Therefore,  he  lias  586 
acres.     All  similar  examples  are  solved  in  liko  manner. 

By  inspecting  the  preceding  illustration,  the  learner 
will  discover  the  following  principle : 

Units  of  the  same  order  are  added  together,  and  the 
sum  is  placed  under  the  column  added.  (Art.  9.) 

NOTES. — i.  The  same  orders  are  placed  under  each  other  for  the 
sake  of  eon  cenience  and  rapidity  in  adding. 

2.  We  add  the  same  orders  together,  units  to  units,  tens  to  tens, 
etc.,  because  different  orders  express  units  of  different  values,  and 
therefore  cannot  be  added  to  each  other.     Thus,  5  units  and  5  tens 
neither  make  10  units  nor  10  tens,  any  more  than  5  cents  and  5 
dimes  will  make  10  cents  or  10  dimes. 

3.  We  add  the  columns  separately,  because  it  is  easier  to  add  one 
order  at  a  time  than  several. 

4.  The  sum  of  each  column  is  set  under  the  column  added,  becaiir  e 
being  less  than  10,  it  is  the  name  order  as  that  column. 

(2.)  (3-)  (4.)  (5-)  (6.) 

2IO3       3O24       I2II       2IO2       2IO32 

4022     1230     2002     I253     52010 

1674      4603      5340      4604      24603 


What  is  the  sum  of  $2321 +$123  +  13245? 


24  ADDITION. 

8.  What  is  the  sum  of  3210  pounds+2023  pounds-f 
4601  pounds? 

9.  What  is  the  sum  of  130230  +  201321-1-402126  ? 
10.  What  is  the  sum  of  24106324-1034246  +  320120  ? 

CASE    II. 

29.  To  find  the  Amount  of  two  or  more  numbers,  when 

the  Sum  of  any  column  is  10,  or  more. 

i.  What  is  the  sum  of  $436,  8324,  and  $645  ? 

ANALYSIS. — Let  the  numbers  be  set  down  as  in  the 
margin.     Adding  as  before,  the  sum  of  the  first  column 
is  15  units,  or  i  ten  and  5  units.     We  set  the  5  units 
under  the  column  added,  and  add  the  i  ten  to  the  next 
column  because  it  is  the  same  order  as  that  columa. 
Now,  i  added  to  4  tens  makes  5  tens,  and  2  are  7  tens,  and 
3  are  10  tens,  or  I  hundred  and  o  tens.    We  set  the  o,  or      $  1 405 
right  hand  figure,  under  the  column  added,  and  add  the 
i  hundred  io  the  next  column,  as  before.     The  sum  of  the  next 
column,  with  the  i  added,  is  14  hundreds  ;  or  I  thousand  and  4  hun- 
dreds.    This  being  the  last  column,  we  set  down  the  wJiote  sum. 
The  answer  is  $1405.     All  similar  examples  may  be  solved  in  like 
manner. 

By  inspecting  this  illustration,  it  will  be  seen, 
When  the  sum  of  a  column  is  i  o  or  more,  we  write  the 
units'  figure  under  the  column,  and  add  the  tens'  figure  to 
the  next  column. 

NOTES. — i.  We  set  the  units'  figure  under  the  column  added,  and 
add  the  tens  to  the  next  column,  because  they  are  the  same  orders 
as  these  columns. 

2.  We  begin  to  add  at  the  right  Juind,  in  order  to  carry  the  tens  as 
we  proceed.  We  set  down  the  whole  sum  of  the  last  column,  because 
there  are  no  figures  of  the  same  order  to  which  its  left  hand  figure 
can  be  added. 

30.  Adding  the  tens  or  left  hand  figure  to  the  next 
column,  is  called  carrying  the  tens.    The  process  of  carry- 
ing the  tens,  it  will  be  observed,  is  simply  taking  a  certain 
number  of  units  from  a  lower  order,  and  adding  their 
equal  to  the  next  higher  ;  therefore,  it  can  neither  increase 
nor  diminish  the  amount. 


m 


ADDITION.  25 

NOTE.— We  carry  for  ten  instead  of  seven,  nine,  eleven,  etc., 
because  in  the  Arabic  notation  the  orders  increase  by  the  scale  of 
ten.  If  they  increased  by  the  scale  of  eight,  twelve,  etc.,  we  should 
carry  for  that  number.  (Art.  13.) 

(*-}  (3-)  (4-)  (5-)  (6.) 

5689  6898  7585  8456  97504 

3792  3365  3748  5078  38786 

4358  7987  8667  6904  75979 

31.  The  preceding  principles  may  be  summed  up 
the  following 

GENERAL  RULE. 

I.  Place  the  numbers  one  under  another,  units  under 
units,  etc. ;  and  beginning  at  the  right,  add  each  column 
separately. 

II.  If  the  sum  of  a  column  does  not  exceed  NINE,  write 
it  under  the  column  added. 

If  the  sum  exceeds  NINE,  write  the  units'  figure  under 
the  column,  and  add  the  tens  to  the  next  higher  order. 
Finally,  set  down  the  whole  sum  of  the  last  column. 

NOTES. — i.  As  soon  as  the  pupil  understands  the  principle  of 
adding,  he  should  learn  to  abbreviate  the  process  by  simply  pro- 
nouncing the  successive  results,  as  he  points  to  each  figure  added. 
Thus,  instead  of  saying  7  units  and  9  units  are  16  units,  and  8  are 
24  units,  and  7  are  31  units,  he  should  say,  nine,  sixteen,  twenty-fair, 
thirty-one,  etc. 

Again,  if  two  or  more  numbers  together  make  10,  as  6  and  4, 
7  and  3  ;  or  2,  3,  and  5,  etc.,  it  is  shorter,  and  therefore  better,  to  add 
10  at  once. 


31.  How  write  numbers  to  be  added  ?  The  next  step  ?  If  the  sum  of  a  column 
does  not  exceed  nine,  what  do  yon  do  with  it  ?  If  it  exceeds  nine  ?  The  sum 
of  the  last  column  ?  28.  Note.  Why  write  units  under  units,  etc.  ?  Why  add  the 
columns  separately  ?  Why  not  add  different  orders  together  promiscuously  ?  Is 
the  sum  of  3  units  and  4  tens,  7  units  or  7  tens  ?  When  the  sum  of  a  column  does 
not  exceed  9,  why  set  it  under  the  column  1  29.  Note.  If  the  sum  of  a  column  is 
10  or  more,  why  set  the  units'  figure  under  the  column  added,  and  carry  the  tens 
to  the  next  column  ?  30.  What  is  meant  by  carrying  the  tens  ?  Why  does  not 
carrying  change  the  amount  f  Why  carry  for  10  instead  of  6,  8,  12,  etc. 


26  ADDITION. 

2.  Accountants  sometimes  set  the  figure  carried  under  the  right 
hand  figure  in  a  line  below  the  answer.  In  this  way  the  sum  of 
each  column  is  preserved,  and  any  part  of  the  work  can  be  reviewed, 
if  desired  ;  or  if  interrupted,  can  be  resumed  at  pleasure. 

32.  PEOOF. — Begin  at  the  top  and  add  each  column 
downward.  If  the  two  results  agree,  the  work  is  right. 

NOTE. — This  proof  depends  upon  the  supposition  that  reversing 
the  order  of  the  figures,  will  detect  any  error  that  may  have  occurred 
in  the  operation. 

EXAMPLES. 

i.  Find  the  sum  of  864,  741,  375,  284,  and  542,  and 
prove  the  operation. 

(2.)  (3-)  (4-)  (5-)  (6.) 

Dollars.  Pounds.  Yards.  Rods.  Feet. 

263  4780  2896  23721  845235 

425  7642  8342  70253  476234 

846  5036  257  4621  6897 

407  7827  3261  342  723468 


7.  What  is  the  sum  of  675  acres +  842  acres +  904  acres 
4-39  acres? 

8.  What  is  the  sum  of  8423  +  286  +  7932  +  28  +  6790? 

9.  What  is  the  sum  of  82431  +  376  +  19  +  62328  +  4521 

+  35787? 
10.  What  is  the  sum  of  63428  +  78  +  4236  +  628  +  93  + 

8413? 

(n.)  (12.)  (I3.)  .(14.)  (15-) 

2685  89243  72094  825276  9°3I253 

6543  8284  96308  704394  432567 

8479  34567  763  37783  65414 

6503  865  4292  1697  9236 

1762  3952  23648  349435  843 

7395  42678  75965  697678  68 

i6.  If  a  man  pays  $2358  for  his  farm,  $1950  for  stock, 
and  $360  for  tools,  how  much  does  he  pay  for  all  ? 

32.  How  prove  addition  ?    Note.  Upon  what  is  this  proof  based  ? 


ADDITION.  27 

17.  A  merchant  bought  371  yards  of  silk,  287  yards  of 
calico,  643  yards  of  muslin,  and  75  yards  of  broadcloth: 
how  many  yards  did  he  buy  in  all  ? 

18.  Eequired  the  sum  of  $2404 +  $100  +  11965 +$1863. 

19.  Eequired  the  sum  of  968  pounds +  81   pounds +  7 
pounds +  639  pounds. 

20.  Required  the   sum  of   1565   gals. +  870  gals. +31 
gals.  + 1 60  gals.  +  42  gals. 

21.  Add  2368,  1764,  942,  87,  6,  and  5271. 

22.  Add  281,  6240,  37,  9,  1923,  101,  and  45. 

23.  Add  888,  9061,  75,  300,  99,  6,  and  243. 

24.  243  +  765+980  +  759  +  127— how  many? 

25.  9423  +  100  + 1600  +  ii9  +  4oo4=how  many? 

26.  81263  + 16319  +  805 +25oo  +  93=how  many? 

27.  236517  +460075  +  2353oo  +  275i6i  =  how  many? 

28.  A  savings  bank  loaned  to  one  customer  $1560,  to 
another  $1973,  to  a  third  $2500,  and  to  a  fourth  $3160: 
how  much  did  it  loan  to  all  ? 

29.  A  young  farmer  raised  763  bushels  of  wheat  the 
first  year,  849  bushels  the  second,  ion  bushels  the  third, 
and  1375  bushels  the  fourth :  how  many  bushels  did  he 
raise  in  4  years  ? 

30.  A  man  bequeathed  the  Soldiers'  Home  $8545 ;  the 
Blind  Asylum   $7538;    the   Deaf   and    Dumb   Asylum 
$6280;  the  Orphan  Asylum  $19260;  and  to  his  wife  the 
remainder,  which  was  $65978.     What  was  the  value  of 
his  estate  ? 

31.  A  merchant  owns  a  store  valued  at  $17265,  his 
goods  on  hand  cost  him  $19230,  and  he  has  $1563  in 
bank :  how  much  is  he  worth  ? 

32.  "What  is  the  sum  of  thirteen  hundred  and  sixty- 
three,  eighty-seven,  one  thousand  and  ninety-four,  and 
three  hundred  ? 

33.  "What  is  the  sum  of  three  thousand  two  hundred 
and  forty,  fifteen  hundred  and  sixty,  and  nine  thousand? 


28  ADDITION. 

34.  Add  ninety  thousand  three  hundred  and  two,  sixty- 
five  thousand  and  thirty,  forty-four  hundred  and  twenty- 
three. 

35.  Add  eight  hundred  thousand  and  eight  hundred, 
forty  thousand  and  forty,  seven  thousand  and  seven,  nine 
hundred  and  nine. 

36.  Add  30006,  301,  55000,  2030,  67,  and  95000. 

37.  Add  65139,  100100,  39,  iniii,  763002,  and  317. 

38.  Add  8 1,  907,  311,  685,  9235,  7,  and  259. 

39.  Add  4895,  352,  68,  7,  95,  645,  and  3867. 

40.  Add  631,  17,  i,  45,  9268,  196,  and  3562. 

41.  Add  77777,  33333>  88888,  22222,  and  nui. 

42.  Add  236578,  125,  687256,  and  4°45°5- 

43.  Add  23246,  8200461,  5017,  8264,  and  39. 

44.  Add  317,  21,  9,  4500,  219,  3001,  17036,  and  45. 

45.  Add  10000000,  joooooo,  100000,  10000,  1000,  100, 
und  10. 

46.  Add  22000000,  22000,  202000,  22oo,  and  2 20. 

47.  If  a  young  man  lays  up  $365  in  i  year,  how  much 
will  he  lay  up  in  4  years  ? 

48.  In  what  year  will  a  man  who  was  born  in  1850,  be  75 
years  old  ? 

49.  A  man  deposits  $1365  in  a  bank  per  day  for  6  days 
in  succession:  what  amount  did  he  deposit  during  the 
week? 

50.  A  planter  raised  1739  pounds  of  cotton  on  ono 
section  of  his  estate,  703  pounds  on  another,  .20 15  pounds 
on  another,  and   2530  pounds  on  another:  how  many 
pounds  did  he  raise  in  all  ? 

51.  A  man  was  29  years  old  when  his  eldest  son  was 
born ;  that  son  died  aged  47,  and  the  father  died  1 7  years 
later :  how  old  was  the  man  at  his  death  ? 

52.  A  speculator  bought  3  city  lots  for  $21213,  an^  sold 
so  as  to  gain  $375  on  each  lot:  for  what  sum  did  he  sell 
them? 


ADDITION.  29 

53.  A's  income-tax  for  1865  was  $4369,  B's  $3978;  C's 
was  $135  more  than  A's  and  B's  together,  and  D's  was 
equal  to  all  the  others:  what  was  D's  tax?    What  the 
tax  of  all  ? 

54.  How  many  strokes  does  a  clock  strike  in  24  hours? 

55.  A  raised  3245  bushels  of  corn,  and  B  raised  723 
bushels  more  than  A :  how  many  bushels  did  both  raise  ? 

56.  In  a  leap  year  7  months  have  31  days  each,  4  months 
30  days  each,  and  i  month  has  29  days:  how  many  days 
constitute  a  leap  year  ? 

57.  The  entire  property  of  a  bankrupt  is  $2648,  which 
is  only  half  of  what  he  owes :  how  much  does  he  owe  ? 

58.  What  number  is  19256  more  than  31273  ? 

59.  A  has  860  acres  of  land,  B  117  acres  more  than  A, 
and  G  as  many  as  both :  how  many  acres  have  all  ? 

DRILL     COLUMNS. 


(60.) 

(61.) 

(62.) 

(630 

(640 

(650 

Pols.  cts. 

Dols.  cts. 

Dols.  cts. 

Dols.  cts. 

Dols.  cts. 

Dols.  cts. 

14  65 

25  76 

37  38 

24  91 

34  45 

49  68 

21  43 

32  37 

3  25 

42  73 

32  61 

26  52 

64  61 

54  61 

46  72 

9  61 

64  12 

38  43 

37  28 

16  45 

28  41 

32  44 

7°  33 

27  59 

43  24 

67  38 

7  5° 

51  62 

24  54 

12  38 

46  03 

34  92 

63  04 

9  28 

32  67 

45  63 

19  41 

76  41 

16  28 

7°  36 

48  32 

5°  7i 

32  34 

47  69 

7  39 

84  52 

25  67 

26  83 

34  32 

3i  04 

82  01 

19  24 

13  09 

34  26 

15  73 

83  26 

7  63 

32  41 

58  32 

72  61 

62  64 

27  13 

24  07 

42  35 

24  63 

23  45 

The  ability  to  add  with  rapidity  and  correctness  is  con- 
fessedly a  most  valuable  attainment ;  yet  it  is  notorious  that  few 
pupils  ever  acquire  it  in  school.  The  cause  is  the  neglect  of  the 
Table,  and  the  want  of  drilling.  Practice  is  the  price  of  skill. 


30  ADDITION. 


(66.) 

(67.) 

(68.) 

(69.) 

(7°.) 

Dols.  cts. 

Dols.  cts. 

Dols.  cts. 

Dols.  cts. 

Dols.  cts. 

25  63 

346  25 

273  42 

3424  27 

4386  48 

46  74 

405  31 

534  97 

6213  39 

3275  26 

82  31 

830  62 

286  31 

5382  13 

5327  64 

60  46 

642  43 

347  34 

4561  63 

4613  04 

75  38 

761  38 

721  35 

8324  29 

1729  56 

22  76 

482  71 

635  26 

5276  34 

3537  63 

64  28 

395  65 

453  94 

6594  S2 

8274  31 

37  31 

762  34 

587  63 

1723  45 

5026  73 

25  16 

25°  25 

658  92 

2674  56 

i586  37 

32  10 

874  5° 

894  28 

3295  31 

2345  67 

75  87 

328  25 

946  25 

4463  79 

4326  43 

62  75 

432  12 

973  22 

5324  28 

5437  26 

48  12 

5i9  75 

564  33 

6543  20 

6325  4i 

23  15 

438  29 

283  12 

7035  28 

7396  27 

18  ii 

533  25 

597  3i 

8546  37 

8325  34 

5°  25 

453  62 

296  54 

9634  25 

9274  S2 

78  26 

374  S2 

458  73 

8346  52 

3465  23 

39  44 

25°  37 

605  42 

6275  35 

4289  67 

72  51 

862  75 

486  54 

4235  34 

5365  72 

33  4i 

953  25 

193  12 

3271  05 

6582  39 

58  75 

846  62 

586  32 

6137  94 

1205  32 

29  31 

553  12 

101  53 

6283  59 

4396  84 

54  62 

228  51 

253  72 

4346  32 

5724  33 

27  3i 

312  52 

XS4  53 

7294  58 

1065  43 

29  50 

455  63 

276  32 

6275  63 

4953  27 

68  71 

729  3i 

586  34 

3284  32 

2586  54 

97  53 

426  76 

235  20 

1635  34 

4234  62 

82  43 

623  25 

463  52 

2586  89 

i736  44 

64  25 

321  35 

958  76 

7434  26 

539s  29 

18  12 

238  17 

386  29 

5869  73 

1234  56 

J9  5° 

125  51 

3*5  46 

3276  42 

7891  01 

62  25 

436  25 

434  57 

1635  38 

1234  16 

64  37 

536  63 

372  46 

59*3  84 

6843  75 

53  63 

257  47 

657  32 

6284  35 

7616  24 

STJBTEAOTION. 

33.  Subtraction  is  taking  one  number  from  another. 
^The  Subtrahend  is  the  number  to  be  subtracted. 

The  Minuend  is  the  number  from  which  the  sub- 
traction is.  made. 

The  Difference  or  Remainder  is  the  number  found 
by  subtraction.  Thus,  when  it  is  said,  5  taken  from  12 
leaves  7,  12  is  the  minuend;  5  the  subtrahend;  and  7  the 
difference  or  remainder. 

NOTES. — i.  The  term  subtraction,  is  from  the  Latin  subtraho,  to 
draw  from  under,  or  take  away. 

The  term  subtrahend,  from  the  same  root,  signifies  that  which  is  to 
be  subtracted. 

Minuend,  from  the  Latin  minuo,  to  diminisJi,  signifies  that  which 
is  to  be  diminished  ;  the  termination  nd  in  each  caso  having  the  force 
of  to  be. 

2.  Subtraction  is  the  opposite  of  Addition.  One  unites,  the  other 
separates  numbers. 

3  When  both  numbers  are  the  same  denomination,  the  operation 
is  called  Simple  Subtraction. 

34.  The  Sign  of  Subtraction  is  a  short  horizontal 
line  called  minus  (  —  },  placed  before  the  number  to  be 
subtracted.     Thus,  6—4  means  that  4  is  to  be  taken  from 
6,  and  is  read,  "  6  minus  4." 

NOTE. — The  term  minus,  Latin,  signifies  less,  or  diminished  by. 
Head  the  following  expressions : 

i.  21—  3  =  2+   3  +  13-         2.     31  —  10=  6+  4+n. 
3.  45  —  15  =  8  +  12  +  10.         4.  100—30  =  55  +  10+   5. 


33.  What  is  subtraction  ?  What  is  the  number  to  be  subtracted  called  ?  The 
number  from  which  the  subtraction  is  made?  The  result?  Note.  Meaning  of 
term  Subtraction  ?  Subtrahend  ?  Minuend  ?  Difference  between  Subtraction 
and  Addition  ?  When  the  Nos.  are  the  same  denomination,  what  ia  the  operation 
called  1  34.  Sign  of  Subtraction  ?  How  read  6  -4  ?  Meaning  of  the  term  minus ; 


SUBTRACTION". 


SUBTRACTION   TABLE. 


I  from 

2  from 

3  from 

4  from 

5  from 

i  leaves  o 

2  leaves  o 

3  leaves  o 

4  leaves  o 

5  leaves  o 

2       «          I 

3     "      i 

4     "       i 

5      "       I 

6     «      i 

3     "      2 

4     "       2 

5     "      2 

6       «         2 

7     «      2 

4     "      3 

5     «      3 

6     «      3 

7      "       3 

8     "      3 

5     "      4 

6     "      4 

7     "      4 

8     «       4 

9     "      4 

6     "      5 

7     "      5 

8     «      5 

9     "      5 

10     «      5 

7     "      6 

8     "      6 

9     "      6 

10     "      6 

ii     «      6 

8     "      7 

9     "      7 

10     "      7 

ii     «      7 

12     "      7 

9     "      8 

10     «      8 

ii     «      8 

12       «         8 

13     "      8 

10     "      9 

ii     "      9 

12     «      9 

13     «      9 

14     "      9 

II       "      10 

12       "      10 

13     «    10 

14     «    10 

15       "      10 

6  from 

7  from 

8  from 

9  from 

10  from 

6  leaves  o 

7  leaves  o 

8  leaves  o 

9  leaves  o 

i  o  leaves  o 

7     "      i 

8     "       I 

9     «      i 

10       "         I 

ii      "       i 

8       "         2 

9     "      2 

10       "         2 

II       "        2 

12       "         2 

9     "      3 

10     «      3 

ii     «      3 

12     "      3 

i3     "      3 

10     "      4 

ii     "      4 

12       "         4 

13     "      4 

14     '*      4 

ii     «      5 

12     "      5 

13     "      5 

14     "      5 

15     «      5 

12       "         6 

13     "      6 

14     "      6 

15     "      6 

16     «      6 

13     "      7 

14    "      7 

15     "      7 

16     «      7 

17     "      7 

14     "      8 

15     "      8 

16     "      8 

17     "      8 

18     «      8 

15     "      9 

16     "      9 

17     «      9 

18     "      9 

19     «      9 

16     «    10 

17     «    10 

18     "    10 

19     "    10 

20       "      10 

It  will  aid  the  pupil  in  learning  the  Subtraction  Table  to 
observe  that  it  is  the  reverse  of  Addition.  Tims,  he  has  learned 
that  3  and  2  are  5.  Reversing  this,  he  will  see  that  2  from  5  leaves 
3,  etc.  Exercises  combining  the  two  tables  will  be  found  useful,  as  a 
review. 

CASE    I. 

35.  To  find  the  Difference, when  each  figure  of  the  Subtra- 
hend is  less  than  the  corresponding  figure  of  the  Minuend. 

Ex.  i.  From  568  dollars,  take  365  dollars. 

ANALYSIS  — Let  the  numbers  be  set  down,  the  less  OPBBATIOK. 
tinder  the  greater,  as  in  the  margin.  Beginning  at  $568  Mln. 
the  right  hand,  we  proceed  thus :  5  units  from  8  units  365  gnl)t> 

leave  3  units;    set  the  3  in  units'  place  under   the     

figure  subtracted,  because  it  is  units.    Next,  6  tens     $203    Rem. 

from  6  tens  leave  o  tens.    Set  the  o  in  tens'  place 

under  the  figure  subtracted,  because  there  are  no  tens.     Finally,  3 


SUBTRACTION.  33 

hundreds  from  5  hundreds  leave  2  hundreds.  Set  the  2  in  hundreds' 
place,  because  it  is  hundreds.  The  remainder  is  $203.  All  similar 
examples  are  solved  in  like  manner. 

By  inspecting  this  illustration,  the  learner  will  discover 
the  following  principle : 

Units  of  the  same  order  are  subtracted  one  from 
another,  and  the  remainder  is  placed  under  the  figure  suo- 
tr acted.  (Arts.  9,  28.) 

NOTES. — 1.  The  less  number  is  placed  under  the  greater,  with 
units  under  units,  etc.,  for  the  sake  of  convenience  and  rapidity 
in  subtracting. 

2.  Units  of  the  same  order  are  subtracted  from  each  other,  for 
the  same  reason  that  they  are  added  to  each  other.  (Art.  28,  n.) 

3.  We  subtract  the  figures  of  the  subtrahend  separately,  because  it 
is  easier  to  subtract  one  figure  at  a  time  than  several. 

4.  The  remainder  is  set  under  the  figure  subtracted,  because  it  is 
the  same  order  as  that  figure. 


From 

(»•) 

648 

(30    (4-)     (50     (60 
726    8652    7230    9621 

Take 

415 

516    3440    412 

o    8510 

(7-) 

(8.) 

(90     (10.).    (ii.) 

From 

53°4 

6546 

7852    8462    9991 

Take 

IIO2 

32I4 

5220    413°    8880 

(12.) 

(130 

(MO 

(isO 

From 

$6948 

43iS  ft- 

$5346 

9657  yds. 

Take 

$2416 

3212  ft. 

$3106 

5021  yds. 

(16.) 

(170 

(18.) 

(190 

From 

645921 

8256072 

72567803 

965235804 

Take 

435010 

6i35052 

42103201 

604132402 

20.  What  is  the  difference  between  3275  and  2132? 

21.  "What  is  the  difference  between  5384  and  3264? 

22.  A  father  gave  his  son  $5750  and  his  daughter  §42505 
how  much  more  did  he  give  his  son  than  his  daughter? 


34  SUBTRACTION. 

CASE     II. 

36.  To  find  the  Difference,  when  a  figure  in  the  Subtra- 
hend is  greater  than  the  corresponding  figure  of  the  Minuend. 

Ex.  i.  What  is  the  difference  between  8074  and  4869  ? 

ist  ANALYSIS. — Let  the  numbers  be  set  down  as  in    ist  METHOD. 
the  margin.      Beginning  at   the  right,   we  proceed     8074    Min. 
thus:    9   units  cannot   be  taken  from  4  units;   we     A 860    Sub 
therefore  take  i  ten  from  the  7  tens  in  the  upper 
number,   and  add  it  to   4   units,  making   14.      Sub-    -120$    Rem. 
tracting  9  units  from  14  units  leaves  5  units,  which 
we  place  under  the  figure  subtracted.     Since  we  have  taken  i  ten 
from  7  tens,  there  are  but  6  tens  left,  and  6  tens  from  6  tens  leaven 

0  tens.     Next,  8  hundreds  cannot  be  taken  from  o  hundreds;  hence, 
we  take  i  thousand  from  the  8  thousand,  and  adding  it  to  the  o, 
we  have  10  hundreds.   Taking  8  hundreds  from  10  hundreds  leaves 
2  hundreds.     Finally,  4  thousand  from  7  thousand  (8 — i)  leave  3 
thousand.     The  remainder  is  3205. 

2d  ANALYSIS. — Adding  10  to  4,  the  upper  figure,  zd  METHOD. 

makes  14,  and  9  from  14  leaves  5.    Now,  to  balance  8074    Min. 

the  ten   added  to  the  upper   figure,   we  add    i   to  4860    Sub 
the  next  higher  order  in  the  lower  number.     Adding 

1  to  6  makes  7,  and  7  from  7  leaves  o.     Next,  since  -j2o";    Rem 
we  cannot  take  8  from  o,  we  add  10  to  the  o,  and  8 

from  10  leaves  2.     Finally,  adding  i  to  4  makes  5,  and  5  from  8 
leaves  3.    The  remainder  is  3205,  the  same  as  before. 

NOTE. — The  chief  difference  between  the  two  methods  is  this :  In 
the  first,  we  subtract  i  from  the  next  figure  in  the  upper  number ;  in 
the  second,  we  add  i  to  the  next  figure  in  the  lower  number.  The 
former  is  perhaps  the  more  philosophical ;  but  the  latter  is  more 
convenient,  and  therefore  generally  practiced. 

By  inspecting  this  illustration,  it  will  be  seen, 

If  a  figure  in  the  lower  number  is  larger  than  that  above 

it,  we  add  10  to  the  upper  figure,  then  subtract,  and  add  i 

to  the  next  figure  in  the  lower  number. 

Hem  — Instead  of  adding  10  to  the  upper  figure,  many  prefer  to 
take  the  lower  figure  directly  from  10,  and  to  the  remainder  add  the 
tipper  figure.  Thus,  9  from  10  leaves  i,  and  4  make  5,  etc. 

37.  Adding  10  to  the  upper  figure  is  called  borrowing ; 
and  adding  i  to  the  next  figure  in  the  lower  number 
pays  it. 


SUBTRACTION.  35 

NOTES. — i.  The  first  method  of  borrowing  depends  upon  the  obvi- 
ous principle  that  the  value  of  a  number  is  not  altered  by  transfer- 
ring a  unit  from  a  higher  order  to  the  next  lower. 

2.  The  reason  that  the  second  method  of  borrowing  does  not  affect 
the  difference  between  the  two  numbers,  is  because  they  are  equally 
increased ;  and  when  two  numbers  are  equally  increased,  their  dif- 
ference is  not  altered. 

3.  The  reasor  for  borrowing  10,  instead  of  5,  8,  12,  or  any  other 
number,  is  because  ten  of  a  lower  order  are  equal  to  one  of  the  nest 
higher.     If  numbers  increased  by  the  scale  of  5,  we  should  add  5  to 
the  upper  figure ;  if  by  the  scale  of  8,  we  should  add  8,  etc.    (Art.  9.) 

4.  We  begin  to  subtract  at  the  right,  because  when  we  borrow  we 
must  pay  before  subtracting  the  next  figure. 

(2.)         (3-)          (4-)  (5-)  (6.) 

From   784     842     6704     8042     9147 
Take    438     695     3598     5439     8638 

38.  The  preceding  principles  may  be  summed  up  in  the 
following 

GENERAL    RULE. 

I.  Place  the  less  number  under  the  greater,  units  under 
units,  etc. 

II.  Begin  at  the  right,  and  subtract  each  figure  in  the 
lower  number  from  the  one  above  it,  setting  the  remainder 
under  the  figure  subtracted.     (Art.  35.) 

III.  If  a  figure  in  the  lower  number  is  larger  than  the 
one  above  it,  add  10  to  the  upper  figure  ;  then  subtract,  and 
add  i  to  the  next  figure  in  the  lower  number.     (Art.  36.) 

39.  PROOF. — Add  the  remainder  to  the  subtrahend ;  if 
the  sum  is  equal  to  the  minuend,  the  work  is  right. 

38.  How  write  numbers  for  subtraction  ?  How  proceed  when  a  figure  in  the 
lower  number  is  greater  than  the  one  above  it?  35.  Note.  Why  write  the  lesa 
number  under  the  greater,  etc.  ?  Why  subtract  the  figures  separately  ?  Why 
set  the  remainder  under  the  figure  subtracted  ?  37.  What  is  meant  by  borrow- 
ing 1  What  is  meant  by  paving  or  carrying  ?  Note.  Upon  what  principle  does 
the  first  method  of  borrowing  depend  ?  Why  does  not  the  second  method  of 
borrowing  affect  the  difference  between  the  two  numbers?  Why  borrow  10 
Instead  of  5,  8,  12,  etc.?  Why  begin  to  subtract  at  the  right  hand?  39.  How 
prove  subtraction  ?  Note.  Upon  what  does  this  proof  depend? 


30  SUBTRACTION. 

NOTES. — i.  This  proof  depends  upon  the  Axiom,  that  the  wJtole  it 
equal  to  the  sum  of  all  its  parts. 

2.  As  soon  as  the  Class  understand  the  rule,  they  should  learn  to 
omit  all  words  but  the  results.  Thus,  ;n  Ex.  2,  instead  of  saying  9 
from  4  you  can't,  9  from  14  leaves  5,  etc.,  sayjice,  naught,  three,  etc. 

EXAMPLES. 
i.  Find  the  difference  between  84065  and  30428. 

(2.)  (3-)  (4-)  (50 

From       824  rods,       4523  pounds,        6841  years,       735°  acresi 
Take        519  rods.       445^  pounds,        3062  years.       5^3*  acre8k 


(6.) 

(7-) 

(8.) 

(9-) 

From  23941 

46083 

52300 

483672 

Take  12367 

23724 

25121 

216030 

(10.) 

(II.) 

(12.) 

(130 

From  638024 

7423614 

8605240 

9042849 

Take  403015 

2414605 

3062431 

6304120 

14.  From  85269  pounds,  take  33280  pounds. 

15.  From  412685  tons,  take  103068  tons. 

'   1 6.  From  840005  acres,  take  630651  acres. 

17.  What  is  the  difference  between  95301  and  60358? 

18.  What  is  the  difference  between  1675236  and  439243  ? 

19.  Subtract  2036573  from  5670378. 

20.  Subtract  35000384  from  68230460. 

21.  Subtract  250600325  from  600230021. 

22.  A  man  bought  a  house  and  lot  for  636250,  and 
paid  $17260  down  :  how  much  does  he  still  owe  ? 

23.  A  man  bought  a  farm  for  $19200,  and  sold  it  for 
$17285  :  what  was  his  loss? 

24.  A  merchant  bought  a  cargo  of  tea  for  $1265235,  and 
Bold  it  for  $1680261  •  what  was  his  gain? 

25.  A's  income  is  §645275,  and  B's  $845280:  what  is 
the  difference  in  their  incomes  ? 


SUBTRACTION.  3? 

26.  A  bankrupt's   assets  are   $569257,  and  his  debts 
$849236  :  how  much  will  his  creditors  lose  ? 

27.  Subtract  9999999  from  iiiiino. 

28.  Subtract  666666666  from  7000000000. 

29.  Subtract  8888888888  from  10000000000. 

30.  Take  200350043010  from  490103060756. 

31.  Take  53100060573604  from  80130645002120. 

32.  Take  675000000364906  from  901638000241036. 

33.  From  two  millions  and  five,  take  ten  thousand. 

34.  From  one  million,  take  forty-five  hundred. 

35.  From  sixty-five  thousand  and  sixty-five,  take  five 
hundred  and  one. 

36.  From  one  billion  and  one  thousand,  take  one  million. 

37.  Our  national  independence  was  declared  in  1776: 
how  many  years  since  ? 

38.  Our  forefathers  landed  at  Plymouth  in  1620:  how 
many  years  since  ? 

39.  A  father  having  3265  acres  of  land,  gave  his  son 
1030  acres:  how  many  acres  has  he  left? 

40.  The  Earth  is  95300000*  miles  from  the  sun,  and 
Venus  68770000  miles:  required  the  difference. 

41.  Washington  died  in  1799,  at  the  age  of  67  years:  in 
what  year  was  he  born  ? 

42.  Dr.  Franklin  was  born  in  1706,  and  died  in  1790: 
at  what  age  did  he  die  ? 

43.  Sir  Isaac  Newton  died  in  1727,  at  the  age  of  85 
years :  in  what  year  was  he  born  ? 

44.  Jupiter  is   496000000  miles   from   the   sun,   and 
Saturn  909000000  miles :  what  is  the  difference  ? 

45.  In  1865,  the  sales  of  A.  T.  Stewart  &  Co.,  by  official 
returns,  were  $39391688;    those  of  H.  B.  Claflin  &  Co., 
$42506715  :  what  was  the  difference  in  their  sales? 

46.  The  population  of  the  United  States  in  1860  was 
31445080;  in  1850  it  was  23191876:  what  was  the  in  crease? 

*  Kiddle's  Astronomy. 

409537 


SUBTRACTION. 


QUESTIONS    FOR    REVIEW. 

1.  A  man's  gross  income  was  $1565  a  month  for  two 
successive  months,  and   his  outgoes  for  the  same  time 
$965  :  what  was  his  net  profit  ? 

2.  A  man  paid  $2635  for  his  farm,  and  $758  for  stock; 
he  then  sold  them  for  $4500 :  what  was  his  gain  ? 

3.  A  flour  dealer  has  3560  barrels  of  flour;  after  selling 
1380  barrels  to  one  customer,  and  985  to  another,  how 
many  barrels  will  he  have  left  ? 

4.  If  I  deposit  in  bank  $6530,  and  give  three  checks  of 
$733  each,  how  much  shall  I  have  left  ? 

5.  What  is  the  difference  between  3658  +  256  +  4236  and 
2430  +  1249? 

6.  What  is  the  difference  between  6035+560  +  75  and 
5003  +  360  +  28? 

7.  What  is  the  difference  between  891+306+5007  and 
40  +  601  +  1703  +  89? 

8.  What  is  the  difference  between  900130  +  23040  and 
19004  +  100607  ? 

9.  A  man  having  $16250,  gained  $3245   by  specula- 
tion, and  spent  $5203  in  traveling:   how  much  had  he 
left? 

10.  A  farmer  bought  3  horses,  for  which  he  gave  $275, 
$320,  and  $418  respectively,  and  paid  $50  down:  how 
much  does  he  still  owe  for  them  ? 

11.  A  young  man    received  three  legacies  of  $3263, 
$5490,  and  $7205  respectively;  he  lost  $4795  minus  $1360 
by  gambling :  how  much  was  he  then  worth  ? 

12.  What  is   the  difference   between   6286  +  850  and 
6286-850? 

13.  What  is  the  difference  between  11325  —  2361  and 
8030-3500?- 

14.  What  number  added  to  103256  will  make  215378? 

15.  What  number  added  to  573020  will  make  700700? 


SUBTBACTIOST.  39 

1 6.  What  number  subtracted  from  230375  will  leave 
121487  ? 

17.  What  number   subtracted  from  317250  will  leave 
190300? 

1 8.  If  the  greater  of  two  numbers  is  59253,  and  the 
difference  is  21231,  what  is  the  less  number? 

19.  If  the  difference  between  two  numbers  is  1363,  and 
the  greater  is  45261,  what  is  the  less  number? 

20.  The  sum  of  two  numbers  is  63270,  and  one  of  them 
is  29385  :  what  is  the  other? 

21.  What  number  increased  by  2343  —  131,  will  become 
3265—291  ? 

22.  What  number  increased  by  3520+1060,  will  become 

6539-279? 

23.  What  number  subtracted  from  5009,  will  become 
2340  +  471? 

24.  Agreed  to  pay  a  carpenter  $5260  for  building  a 
house;  $3520  for  the  masonry,  and  $1950  for  painting: 
how  much  shall  I  owe  him  after  paying  $6000  ? 

25.  If  a  man  earns  $150  a  month,  and  it  costs  him  $63 
a   month    to    support    his    family,   how   much  will    he 
accumulate  in  6  months  ? 

26.  A's  income-tax  is  $1165,  B's  is  $163  less  than  A's; 
and  C's  is  as  much  as  A  and  B's  together,  minus  $365 : 
what  is  C's  tax  ? 

27.  A  is  worth  $15230,  B  is  worth  $1260  less  than  A; 
and  C  is  worth  as  much  as  both,  wanting  $1760:  what  is 
B  worth,  and  what  C  ? 

28.  The  sum  of  4  numbers  is  45260;  the  first  is  21000, 
the  second  8200  less  than  the  first,  the  third  7013  less 
than  the  second  :  what  is  the  fourth? 

29.  A  sailor  boastingly  said;    If  I    could   save  $26.3 
more,  I  should  have  $1000:  how  much  had  he  ? 

30.  The  difference  between  A  and  B's  estates  is  61525  ; 
B,  who  has  the  least,  is  worth  $17250:  what  is  A  worth  ? 


MULTIPLICATION. 

40.  Multiplication  is  finding  the  amount  of  a 

number  taken  or  added  to  itself,  a  given  number  of  times. 

The  Multiplicand  is  the  number  to  be  multiplied. 

The  Multiplier  is  the  number  by  which  we  mul- 
tiply. 

The  Product  is  the  number  found  by  multiplication. 
Thus,  when  it  is  said  that  3  times  6  are  18,  6  is  the  mul- 
tiplicand, 3  the  multiplier,  and  18  the  product. 

41.  The  multiplier  shows  how  many  times  the  multi- 
plicand is  to  be  taken.     Thus, 

Multiplying  by  i  is  taking  the  multiplicand  once  ; 
Multiplying  by  2  is  taking  the  multiplicand  twice  ;  and 
Multiplying  by  any  whole  number  is  taking  the  multi- 
plicand as  many  times  as  there  are  units  in  the  multiplier. 


NOTES.  —  i.  The  term  multiplication,  from  the  Latin 
multus,  many,  and  plico,  to  fold,  primarily  signifies  to  increase  by 
regular  accessions. 

2.  Multiplicand,  from  the  same  root,  signifies  that  which  is  to  be  mid- 
tiplied  ;  the  termination  nd,  having  the  force  of  to  be.  (Art,  33.  n.) 

42.  The  multiplier  and  multiplicand  are  also  called 
Factors  ;  because  they  make  or  produce  the  product' 
Thus,  2  and  7  are  the  factors  of  14. 

NOTES.  —  i.  The  term  factor,  is  from  the  Latin  facio,  to  produce. 
2.  When  the  multiplicand  contains  only  one  denomination,  the 
operation  is  called  Simple  Multiplication. 

40.  What  is  multiplication?  What  is  the  number  multiplied  called?  The 
number  to  multiply  by?  The  result?  When  it  is  said,  3  times  4  are  12,  which  is 
the  multiplicand?  The  multiplier?  The  product?  41.  What  does  the  multi- 
plier show  ?  What  is  it  to  multiply  by  i  ?  By  2  ?  42.  What  else  aro  (he  nntnbers 
to  be  multiplied  together  called?  Why?  Not'.  Meaning  of  the  term  factor? 
What  is  the  operation  called  when  the  multiplicand  contains  only  one  denomi- 
nation ?  43.  The  sign  of  multiplication. 


MULTIPLICATION. 


43.  The  Sign  of  Multiplication  is  an  oblique 
cross  (  x  ),  placed  between  the  factors.     Thus,  7x5  means 
that  7  and  5  are  to  be  multiplied  together,  and  is  read 
"  7  times  5,"  "  7  multiplied  by  5,"  or  "  7  into  5." 

44.  Numbers  subject  to  the  same  operation  are  placed 
within  &  parenthesis  ( ),  or  under  a  line  called  a  vincuhim 

( ).    Thus  (4  +  5)  x  3,  or  4  +  5  x  3,  shows  that  the  sum 

of  4  and  5  is  to  be  multiplied  by  3. 

MULTIPLICATION  TABLE. 


2  times 

3  times 

4  times 

5  times 

6  times 

7  times 

i  are  2 

i  are  3 

i  are  4 

i  are  5 

i  are  6 

i  are  7 

2  "  4 

2  "   6 

2  "   8 

2  "  10 

2  "  12 

2  "  14 

3  "  6 

3  "  9 

3  «  12 

3  "  15 

3  "  18 

3  «  27 

4  «  8 

4  "  12 

4  "  16 

4  "  20 

4  "  24 

4  "  28 

5  "  10 

5  "  15 

5  "  20 

5  "  25 

5  "  30 

5  "  35 

6  "  12 

6  «  18 

6  "  24 

6  «  30 

6  «  36 

6  «  42 

7  "  i4 

7  "  21 

7  "  28 

7  "  35 

7  «  42 

7  "  49 

8  «  16 

8  "  24 

8  "  32 

8  "  40 

8  «  48 

8  "  56 

9  "  18 

9  "  27 

9  "  36 

9  "  45 

9  "  54 

9  "  63 

10  "  20 

10  «  30 

10  "  40 

10  "  50 

10  «  60 

10  "  70 

II  "  22 

ii  "  33 

ii  "  44 

ii  «  55 

ii  «  66 

n  «  77 

12  "  24 

12  «  36 

12  «  48 

12  "  60  12  "  72 

12  "  84 

8  times 

9  times 

10  times 

ii  times 

12  times 

i  are  8 

i  are  9 

i  are  10 

i  are  n 

i  are  12 

2   «   16 

2  "   l8 

2  "   20 

2  "   22 

2  «   24 

3  *  24 

3  "  27 

3  "  30 

3  "  33 

3  "  36 

4  "  32 

4  "  36 

4  "  40 

4  "  44 

4  "  48 

5  "  40 

5  "  45 

5  "  50 

5  "  55 

5  "  60 

6  "  48 

6  "  54 

6  "  60 

6  «  66 

6  "  72 

7  "  56 

7  «  63 

7  "  7o 

7  "  77 

7  "  84 

8  «  64 

8  "  72 

8  «  80 

8  "  88 

8  «  96 

9  «  72 

9  "  81 

9  «  90 

9  «  99 

9  «  108 

10   "   80 

10  "  90 

IO  "  IOO 

10  "  IIO 

10  "  120 

ii  "  88 

ii  "  99 

II   "lIO 

II  "  121 

II  "  132 

12   "   96 

12  «  108 

12  "  120 

12  "  132 

12  "  144 

Tlie  pupil  will  observe  that  the  products  by  5,  terminate  in 
5  and  o,  alternately.    Thus,  5  times  i  are  5 ;  5  times  2  are  10. 


42  MULTIPLICATION. 

The  products  by  10  consist  of  the  figure  multiplied  and  a  ciplwr. 
Thus,  10  times  i  are  10 ;  ten  times  2  are  20,  and  so  on. 

The  first  nine  products  by  n  are  formed  by  repeating  the  figure 
multiplied.  Thus,  n  times  I  are  n  ;  n  times  2  are  22,  and  so  on. 

The  first  figure  of  the  first  nine  products  by  9  increases,  and  the 
second  decreases  regularly  by  I ;  while  the  sum  of  the  digits  in  each 
product  is  9.  Thus,  9  times  2  are  18,  9  times  3  are  27,  and  so  on. 

45.  When  the  factors  are  abstract  numbers,  8  at  &  S 
it  is  immaterial  in  what  order  they  are  multi-  &  $  $  a 
plied.    Thus,  the  product  of  3  times  4  is  equal  gj  $  0  & 
to  4  times  3.    For,  taking  4  units  3  times,  is 

the  same  as  taking  3  units  4  times ;  that  is,  4  x  3=3  x  4. 

NOTES. — i.  It  is  more  convenient  and  therefore  customary,  to 
take  the  larger  of  the  two  given  numbers  for  the  multiplicand. 
Thus,  it  is  better  to  multiply  5276  by  8,  than  8  by  5276. 

2.  Multiplication  is  the  same  in  principle  as  Addition;  and  is 
sometimes  said  to  be  a  short  method  of  adding  a  number  to  itself  a 
given  number  of  times.  Thus:  4  stars +  4  stars +  4  stars=3  times  4 
or  12  stars. 

46.  The  product  is  the  same  name  or  kind  as  the  multi- 
plicand.    For,  repeating  a  number  does  not  change  its 
nature.  Thus,  if  we  repeat  dollars,  they  are  still  dollars,  etc. 

47.  TJhb  multiplier  must  be  an  abstract  number,  or  con- 
sidered as  such  for  the  time  being.    For,  the  multiplier 
shows  how  many  times  the  multiplicand  is  to  be  taken ;  and 
to  say  that  one  quantity  is  taken  as  many  times  as  another 
\s  heavy — is  obi  urd. 

NOTE. — When  it  is  asked  what  25  cts.  multiplied  by  25  cts.,  or 
23.  6d.  by  2s.  6d.,  will  produce,  the  questions,  if  taken  literally, 
are  nonsense.  For,  2s.  6d.  cannot  be  repeated  2s.  6d.  times,  nor  25  cts. 
25  cts.  times ;  but  we  can  multiply  25  cts.  by  the  number  25,  and 
23.  6d.  by  2^.  In  like  manner  we  can  multiply  the  price  of  i  yard  by 
the  number  of  yards  in  an  article,  and  the  product  will  be  the  cost. 

45.  When  the  factors  are  abstract,  docs  it  make  any  difference  with  the  pro- 
duct which  is  taken  for  the  multiplicand  f  Note.  Which  is  commonly  taken  f 
Why  ?  What  is  Multiplication  the  same  as  ?  46.  What  denomination  is  the  pr« 
duct  ?  Why  ?  47.  What  must  the  multiplier  be  ?  Why  ? 


MULTIPLICATION.  43 

CASE    I. 

48.  To  find  the  Product  of  two   numbers,  when   the 

Multiplier  has  but  one  figure. 

i.  What  is  the  cost  of  3  horses,  at  $286  apiece  ? 

ANALYSIS. — 3  horses  will  cost  3  times  as  much  as  OPEBATION. 

1  horse.     Let  the  numbers  be  set  down  as  in  the  $286  Multd. 
margin.     Beginning  at  the  right,  we  proceed  thus :  ^  Mult. 

3  times  6  units  are  18  units.     We  set  the  8  in  units'  

place  because  it  is  units,  and  carry  the  i  ten  to  the  $858  Prod, 
product  of  the  tens,  as  in  Addition.  (Art.  29.)    Next, 

3  times  8  tens  are  24  tens  and  i  (to  carry)  make  25  tens,  or  2  hun- 
dred and  5  tens.  We  set  the  5  in  tens'  place  because  it  is  tens,  and 
carry  the  2  hundred  to  the  product  of  hundreds.  Finally,  3  times 

2  hundred  are  6  hundred,  and  2  (to  carry)  are  8  hundred.       We  set 
the  8  in  hundreds'  place  because  it  is  hundreds.     Therefore  the  3 
horses  cost  $858.    All  similar  examples  are  solved  in  like  manner. 

By  inspecting  the  preceding  analysis,  the  learner  will 
discover  the  following  principle : 

Each  figure  of  the  multiplicand  is  multiplied  ly  the 
multiplier,  beginning  at  the  right,  and  the  result  is  set 
down  as  in  Addition.  (Art.  29.) 

NOTES. — i.  The  reason  for  setting  the  multiplier  under  the  multi- 
plicand is  simply  for  convenience  in  multiplying. 

2.  The  reasons  for  beginning  to  multiply  at  the  right  hand,  as 
well  as  for  setting  down  the  units  and  carrying  the  tens,  are  the 
same  as  those  in  Addition.     (Art.  29,  TO.) 

3.  In  reciting  the  following  examples,  the  pupil  should  care- 
fully analyze  each  ;  then  give  the  results  of  the  operations  re- 
quired. 

49.  Units  multiplied  by  units,  it  should  be  observed, 
produce  units ;  tens  by  units,  produce  tens ;  hundreds  by 
units,  produce  hundreds  ;  and,  universally, 

If  the  multiplying  figure  is  units,  the  product  will  be 
the  same  order  as  the  figure  multiplied. 
Again,  if  the  figure  multiplied  is  units,  the  product  will 

49.  What  do  units  multiplied  by  units  produce?  Tens  by  units?  Hundredi 
V  units  ?  When  the  multiplying  figure  is  units,  what  is  the  product? 


44  MULTIPLICATION. 

be  the  same  order  as  the  multiplying  figure;  for  the 
product  is  the  same,  whichever  factor  is  taken  for  the 
multiplier.  (Art.  45.) 

('•)  (3-)  (4-)  (50 

Multiply    2563  13278  2648203  48033265 

By  4  5  6 


6.  "What  will  8  carriages  cost,  at  $750  apiece? 

7.  What  cost  9  village  lots,  at  $1375  a  lot? 

8.  At  $3500  apiece,  what  will  10  houses  cost? 

9.  At  $865   a  hundred,  how  much  will    u  hundred- 
weight of  opium  come  to  ? 

10.  If  a  steamship  sail  358  miles  per  day,  how  far  will 
she  sail  in  1 2  days  ? 

11.  Multiply  86504  by  5.        12.  Multiply  803127  by  7. 
13.  Multiply  440210  by  6.      14.  Multiply  920032  by  8. 
15.  Multiply  603050  by  9.      16.  Multiply  810305  by  10. 
17.  Multiply  753825  by  ii.     18.  Multiply  954635  by  12. 

19.  What  cost  4236  barrels  of  apples,  at  $7  a  barrel  ? 

20.  What  cost  5167  melons,  at  n  cents  apiece? 

21.  At  6  shillings  apiece,  what  will  1595  arithmetics 
cost? 

22.  At  $12  a  barrel,  what  will  be  the  cost  of  1350  bar- 
rels of  flour  ? 

23.  What  will  1735  boxes  of  soap  cost,  at  $9  a  box  ? 

24.  When  peaches  are  12  shillings  a  basket,  what  will 
2363  baskets  cost  ? 

25.  If  a  man  travels  8  miles  an  hour,  how  far  will  he 
travel  in  3260  hours  ? 

26.  A  builder  sold  10  houses  for  $4560  apiece:   how 
much  did  he  receive  for  them  all  ? 

27.  What  will  it  cost  to  construct  ii  miles  of  railroad, 
at  $12250  per  mile? 

28.  What  will  be  the  expense  of  building  12  ferry-boats, 
at  $23250  apiece  ? 


MULTIPLICATION.  45 


CASE    II. 

50.     To    find    the    Product   of   two    numbers,    when    the 
Multiplier  has  two  or  more  figures. 

i.  A  manufacturer  bought  204  bales  of  cotton,  at  $176 
a  bale :  what  did  he  pay  for  the  cotton  ? 

ANALYSIS. — Write  the  numbers  as  in  the  mar-        $176  Multd. 
gin.    Beginning  at  the  right :  4  times  6  units  are  2Q.  Mult 

24  units,  or  2  tens  and  4  units.     Set  the  4  in  units'         — • 

place,  and  carry  the  2  to  the  next  product.    4  7°4 

times  7  tens  are  28  tens  and  2  are  30  tens,  or  3      352 

hundred  and  o  tens.     Set  the  o  in  tens'  place  and  

carry  the  3  to  the  next  product.    4  times  i  hun-    $35904  Am. 
are  4  hun.  and  3  are  7  hun.   Set  the  7  jn  hundreds' 
place .    The  product  by  o  tens  is  o ;  we  therefore  omit  it.    Again, 
2  hundred  times  6  units  are  12  hun.  units,  equal  to  i  thousand  and 
2  hundred.      Set  the  2  in  hundreds'  place  and  carry  the   I  to  the 
next  product.     2  hundred  times  7  tens  are  14  hundreds  of  tens,  equal 
to  14  thousand,  and  i  are  15  thousand.     Set  the  5  in   thousands' 
place,  and  carry  the  i  to  the  next  product,  and  so  on.      Adding 
these  results,  the  sum  is  the  cost. 

REMAKK. — The  results  which  arise  from  multiplying  the  multipli- 
cand by  the  separate  figures  of  the  multiplier,  are  called  partial  pro- 
ducts ;  because  they  are  parts  of  the  whole  product. 

By  inspecting  this  analysis,  the  learner  will  discover, 

1.  The  multiplicand  is  multiplied  by  each  figure  of  the 
multiplier,  beginning  at  the  right,  and  the  result  set  down 
as  in  Addition. 

2.  The  first  figure  of  each  partial  product  is  placed 
under  the  multiplying  figure,  and  the  sum  of  the  partial 
products  is  the  true  answer. 

NOTES. — i.  When  there  are  ciphers  between  the  significant 
figures  of  the  multiplier,  omit  them,  and  multiply  by  the  next  sig- 
nificant figure. 

2.  We  multiply  by  each  figure  of  the  multiplier  separately,  when  it 
exceeds  12,  for  the  obvious  reason,  that  it  is  not  convenient  to  multi- 
ply by  the  whole  of  a  large  number  at  once. 

3.  The  first  figure  of  each  partial  product  is  placed  under  the 
multiplying  figure;  because  it  is  the  same  order  as  that  figure. 

4.  The  several  partial  products  are  added  together,  because  tha 
whole  product  is  equal  to  the  sain  of  all  its  parts. 


46  MULTIPLICATION. 

('•)              (30               (4-)  (50 

Multiply    426                563                 1248  2506 

By             24               35                 52  304 


51.  The  preceding  principles  may  be  summed  up  in 
the  following 

GENERAL    RULE 

I.  Place  the  multiplier  under  the  multiplicand)  units 
under  units,  etc. 

II.  When  the  multiplier  has  but  one  figure,  beginning  at 
the  right,  multiply  each  figure  of  the  multiplicand  by  it, 
and  set  down  the  result  as  in  Addition. 

III.  If  the  multiplier  has  two  or  more  figures,  multiply 
the  multiplicand  by  each  figure  of  the  multiplier  separately, 
and  set  the  first  figure  of  each  partial  product  under  the 
multiplying  figure. 

Finally,  the  sum  of  the  partial  products  will  be  the 
answer  required. 

NOTE. — The  pupil  should  early  learn  to  abbreviate  the  several 
steps  in  multiplying  as  in  Addition.  (Art.  31,  n.) 

PROOF. 

52.  By  Multiplication. — Multiply  the  multiplier  by  the 
multiplicand ;  if  this  result  agrees  with  the  first,  the  work 
is  right. 

NOTE. — This  proof  is  based  upon  the  principle  that  the  result 
will  be  the  same,  whichever  number  is  taken  as  the  multiplicand. 

53.  By  excess  of  95. — Find  the  excess  of  9.*?  in  each  factor 
separately ;  then  multiply  these  excesses  together,  and  reject 
the  gs  from  the  result ;  if  this  excess,  agrees  with  the  excess 
of  93  in  the  answer,  the  work  is  right. 

51.  How  write  numbers  for  multiplication?  Wlien  the  multiplier  has  but  one 
figure,  how  proceed  ?  When  it  has  two  or  more  ?  What  is  finally  done  with  the 
partial  products  ?  4*?.  Note.  Why  write  the  multiplier  under  the  multiplicand, 
units  under  units  ?  Why  begin  at  the  right  hand  ?  50.  What  are  partial  products  ? 
Note.  Why  multiply  by  each  figure  separately  ?  Why  set  the  first  figure  of  each 
under  the  multiplying  figure  f  Why  add  the  several  partial  products  together  T 
52.  How  prove  multiplication  by  multiplication  T  Note.  Upon  what  is  this  proof 
based  ?  53.  How  prove  it  by  excess  of  98  f 


MULT  IPLICATIOX. 


47 


NOTE. — This  method  of  proof,  if  deemed  advisable,  may  be  omitted 
till  review.  It  is  placed  here  for  convenience  of  reference.  Though 
depending  on  a  peculiar  property  of  numbers,  it  is  easily  applied, 
and  is  confessedly  the  most  expediitous  method  of  proving  multipli- 
cation yet  devised. 

54.  To  find  the  Excess  of  9s  in  a  number. 

Beginning  at  the  left  hand,  add  the  figures  together,  and  as  soon 
as  the  sum  is  9  or  more,  reject  9  and  add  the  remainder  to  the  next 
figure,  and  so  on. 

Let  it  be  required  to  find  the  excess  of  93  in  7548467. 

Adding  7  to  5,  the  sum  is  12.  Rejecting  9  from  12,  leaves  3  ;  and 
3  added  to  4  are  7,  and  8  are  15.  Rejecting  9  from  15,  leaves  6 ;  and 
<>  added  to  4  are  10.  Rejecting  9  from  10,  leaves  i ;  and  i  added  to 
6  are  7,  and  7  are  14.  Finally,  rejecting  9  from  14  leaves  5,  the 
excess  required. 

NOTES. — i.  It  will  be  observed  that  the  excess  of  93  in  any  two 
digits  is  always  equal  to  the  sum,  or  the  excess  in  the  sum,  of  those 
digits.  Thus,  in  15  the  excess  is  6,  and  1  +  5=6;  so  in  51  it  is  6, 
and  5  +  i=T>.  In  56  the  sum  is  n,  the  excess  2. 

2.  The  operation  of  finding  the  excess  of  93  in  a  number  is  called 
casting  out  the  gs. 

EXAMPLES. 

i.  What  is  the  product  of  746  multiplied  by  475  ? 


OPERATION.                   Proof  by  Excess  of  9*. 

746      Excess  of  93  in  multd.  is  8 
475             «          9s  in  mult.    "  7 

Proof  by  Mult. 

475 
746 

Now  8x7—56 

-222        The  excess  of  93  in  56  is  2 
2984         The  excess  of  93  in  prod,  is  2 

2850 
1900 
3325 

dns.  35435° 
(2.)              (30              (4-) 

Multiply    5645              18934             48367 
By              43                 65                 75 

Ans.  35435° 

(50 
23I456 
87 

54.  How  find  the  excess  of  98  ? 


48  MULTIPLICATION. 

(6.)  (7.)  (8.)  (9.) 

Multiply  1421673    2342678    4392460    5230648 

By  234  402  347  526 

10.  Mult.  640231  by  205.  ii.  Mult.  520608  by  675. 

12.  Mult.  431220  by  1234.  13.  Mult  623075  by  2650. 

14.  Mult.  730650  by  2167.  15.  Mult.  593287  by  6007. 

16.  Mult.  843700  by  3465.  17.  Mult.  748643  by  2100. 

18.  Mult.  9000401  by  5000 1.  19.  Mult.  82030405  by  23456. 

20.  How  many  pounds  in  1375  chests  of  tea,  each  chest 
containing  63  pounds  ? 

21.  What  cost  738  carts,  at  $75  apiece? 

22.  At  43  bushels  per  acre,  how  many  bushels  of  wheat 
will  520  acres  produce? 

23.  At  $163  apiece,  what  will  be  the  cost  of  1368  covered 
buggies  ? 

24.  If  a  man  travel  215  miles  per  day,  how  far  can  he 
travel  in  365  days  ? 

25.  There  are  5280  feet  in  a  mile:  how  many  feet  in 
256  miles? 

26.  What  cost  2115  revolvers,  at  $23  apiece? 

27.  Bought  1978  barrels  of  pickles,  at  $17  ;  how  much 
did  they  come  to  ? 

28.  If  railroad  cars  are  $4735  apiece,  what  will  be  the 
expense  of  500? 

29.  How  far  will  2163  spools  of  thread  extend,  each 
containing  25  yards  ? 

30.  Bought  15265  ambulances,  at.  $117  apiece:  what 
was  the  amount  of  the  bill  ? 

31.  What  will  3563  tons  of  railroad  iron  cost,  at  $68 
per  ton  ? 

32.  How  far  will  a  man  skate  in  6  days,  allowing  he 
skates  8  hours  a  day,  and  goes  )  miles  an  hour  ? 


MULTIPLICATION.  4S 

CONTRACTIONS. 

55.  A  Composite  Number  is  the  product  of  two 
or  more  factors,  each  of  which  is  greater  than  i.  Thus 
15  =  3x5,  and  42  =  2  x  3  x  7,  are  composite  numbers. 

NOTES. — i.  The  product  of  a  number  multiplied  into  Uself  is 
called  a  power.  Thus,  9=3  x  3,  is  a  power. 

2.  Composite  is  from  the  Latin  compono,  to  place  together. 

3.  The  terms  factors  &n&  parts  must  not  be  confounded  with  each 
other.    The  former  are  multiplied  together  to  produce  a  number ; 
rl:e  latter  are  added.    Thus,  3  and  5  are  the  factors  of  15 ;  but  5  and 
10,  6  and  9,  7  and  8,  etc.,  are  the  parts  of  15. 

1.  What  are  the  factors  of  45  ?     Ans.  5  and  9. 

2.  What  are  the  factors  of  24  ?    Of  2 7  ?    Of  28  ?  Of  30  ? 

3.  What  are  the  factors  of  32?    Of  35  ?    Of  42?  Of  48? 

4.  What  are  the  factors  of  54  ?    Of  63?    Of  72?  Of  84? 


CASE   I. 
56.  To  multiply  by  a  Composite  Number. 

i.  A  farmer  sold  i5»boxes  of  butter,  each  weighing  20 
pounds:  how  many  pounds  did  he  sell? 

ANALYSIS. — 15  =  5  times  3;  hence,  15  boxes 

OPERATION. 

will  weigh  5  times  as  much  as  3  boxes.    Now, 

if  i  box  weighs  20  pounds,  3  boxes  will  weigh 

3  times  20,  or  60  pounds.    Again,  if  3  boxes 

weigh  60  pounds,  5  times  3  boxes  will  weigh  5 

times  60,  or  300  pounds.    He  therefore  sold  300 

rounds.    In  the  operation,  we  first  multiply  by  5 

the  factor  3,  and  the  product  thus  arising  by  the 

other  factor  5.    Hence,  the  Ans.  300  pounds, 

EULE. — Multiply  the  multiplicand  by  one  of  (he  facers 
of  the  multiplier,  then  this  product  by  another,  and  so  on, 
till  all  the  factors  have  been  used. 

The  last  product  will  be  the  answer. 

35.  A  composite  number  ?  Note.  The  difference  between  factors  and  pane  t 
?6.  How  multiply  by  a  composite  number  ? 

3 


50  MULTIPLICATION. 

NOTES. — i.  This  rule  is  based  upon  the  principle  that  it  is 
immaterial  in  what  order  two  factors  are  multiplied.  (Art.  45.) 

The  sanifc,  illustration  may  be  extended  to  three  or  more  numbers. 
For,  the  product  of  two  of  the  factors  may  be  considered  as  one 
Dumber,  and  this  may  be  used  before  or  after  a  third  factor,  etc. 

2.  The  process  of  multiplying  three  or  more  factors  together,  ia 
called  Continued  Multiplication;  the  result,  the  continued  product. 

2.  Multiply  $568  by  35.    Ans.  $19880. 

3.  Multiply  2604  by  25.        4.  Multiply  6052  by  48. 
5.  Multiply  8091  by  63.        6.  Multiply  45321  by  72. 

7.  In  i  cubic  foot  there  are  1728  cubic  inches:    how 
many  cubic  inches  are  there  in  84  cubic  feet  ? 

8.  If  a  ton  of  copper  ore  is  worth  $5268,  what  is  the 
worth  of  56  tons? 

9.  What  cost  125  houses,  at  $1580  apiece? 

CASE   II. 

57.  To  multiply  by  1O,  1OO,  1OOO,  etc. 

10.  What  will  100  cows  cost,  at  $31  apiece? 

ANALYSIS. — Annexing  a  cipher  to  a  number  moves  each  figure  one 
place  to  the  left ;  but  moving  a  figure  orie  place  to  the  left  increases 
vis  value  ten  times ;  therefore  annexing  a  cipher  to  a  number 
multiplies  it  by  10.  In  like  manner,  annexing  two  ciphers,  multi- 
plies it  by  a  hundred,  etc.  (Art.  12.)  Therefore,  $31  x  ioo=$3ioo. 
Hence,  the 

RULE. — Annex  as  many  ciphers  to  the  multiplicand  as 
there  are  ciphers  in  the  multiplier. 

NOTE. — The  term  annex,  xrom  the  Latin  ad  and  necto,  to  join  to, 
signifies  to  place  after. 

11.  What  cost  1000  horses,  at  $356  apiece? 

12.  Multiply  40530  by  1000. 

13.  Multiply  9850685  by  10000. 

14.  Multiply  84050071  by  100000. 

15.  Multiply  360753429  by  1000000. 

57.  How  multiply  by  10,  100,  1000.  etc.  ? 


•<2^L 

MULTIPLICATION.  51 


CASE    III. 

58.  To  multiply,  when  there  are  Ciphers  on  the  right  of 
either  or  both  Factors. 

20.  What  is  the  product  of  87000  multiplied  by  230  ? 

ANALYSIS. — The  factors  of  the  multiplicand  are 
87  and  1000;  the  factors  of  the  multiplier  are  23 
and  10.  We  first  multiply  the  factors  consisting 
of  significant  figures  ;  then  multiply  this  product 
by  the  other  two  factors  (1000  x  10),  or  10000, 
by  annexing  4  ciphers  to  it.  Hence,  the  Ans.  20010000 

RULE. — Multiply  the  significant  figures  together ;  and 
to  the  result  annex  as  many  -ciphers  as  are  found  on  the 
right  of  both  factors. 

NOTE. — This  rule  is  based  upon  the  two  preceding  cases;  for,  by 
supposition,  one  or  both  the  given  numbers  are  composite ;  and  one 
of  the  factors  of  this  composite  number  is  10,  100,  etc. 

(21.)          (22.)          (23.) 
Multiply  2130  64000          83046 

By        700          52  2000 


24.  In  i  barrel  of  pork  there  are  200  pounds:  how  many 
pounds  in  3700  barrels? 

25.  What  will  2300  head  of  cattle  cost,  at  $80  per  head? 

26.  In  $i  there  are  100  cts. :  how  many  cts.  in  $26000? 

27.  The  salary  of  the  President  is  $50000  a  year:  how 
much  will  it  amount  to  in  21  years  ? 

28.  If  a  clock  ticks  86400  times  in  i  day,  how  many 
times  will  it  tick  in  7000  days? 

29.  If  one  ship  costs  $150000,  what  will  49  cost? 

30.  670103700  x  60030040? 

31.  800021000x80002100? 
32-  570305000x40000620? 

58.  When  one  or  both  factors  have  ciphers  on  the  right?    Note,  Upon  what  la 
this  rule  based  ? 


52  MULTIPLICATION. 

33.  467234630X27000000? 

34.  890000000x350741237? 

35.  9400000027  x  28000000  ? 

36.  Multiply  39   millions  and   200  thousand  by  530 
thousand. 

37.  Multiply  102  times  700  thousand  and  i  hundred  by 
60 1  thousand  and  twenty. 

38.  Multiply  74  millions  and  21  thousand  by  5  millions 
and  5  thousand. 

39.  Multiply  31  millions   31  thousand  and  31  by  21 
thousand  and  twenty-one. 

40.  Multiply  2  billions,  2  millions,  2  thousand  and  2 
hundred  by  200  thousand  and  2  hundred. 

CASE   IV. 

59.  To  multiply  13,  14,  15,  OP  I  with  a  Significant  Figure 
annexed. 

41.  If  one  city  lot  costs  $3245,  what  will  17  lots  cost? 

ANALYSIS. — 17  lots  will  cost  17  times  as  much  as  3245  x  17 

I  lot.      Placing  the  multiplier   on    the   right,  we  22715 

multiply  the  multiplicand  by  the  7  unite,  set  each  . 

figure  one  place  to  the  right  of  the  figure  multi-  $55165  An. 
plied,  and  add  the   partial   product  to  the  multi- 
plicand.   The  result  is  $55165.    Hence,  the 

EULE. — I.  Multiply  the  multiplicand  by  the  unit*?  figure 
of  the  multiplier,  and  set  each  figure  of  the  partial  product 
one  place  to  the  right  of  the  figure  multiplied. 

II.  Add  tliis  partial  product  to  the  multiplicand,  and  the 
result  will  be  the  true  product. 

NOTE. — This  contraction  depends  upon  the  principle  that  as  the 
tens'  figure  of  the  multiplier  is  i,  the  multiplicand  is  the  second  par- 
tial product ;  hence  its  first  figure  must  stand  in  tens'  place. 

42.  Multiply    1368  by  13.        43.  Multiply    2106  by  14. 
44.  Multiply    3065  by  15.        45.  Multiply    6742  by  16. 
46.  Multiply  25269  by  18.       47.  Multiply  83467  by  19. 

59.  How  multiply  by  13, 14,  15,  or  i  with  a  significant  figure  annexed  ?  Note. 
Upon  what  is  this  contraction  based  ? 


PIVISION. 

60.  Division  is  finding  how  many  times  one  numb**1 
is  contained  in  another. 

The  Dividend  is  the  number  to  be  divided. 

The  Divisor  is  the  number  by  which  we  divide. 

The  Quotient  is  the  number  found  by  division,  and 
shows  how  many  times  the  divisor  is  contained  in  the 
dividend. 

The  Remainder  is  a  part  of  the  dividend  left  after 
division.  Thus,  when  it  is  said,  5  is  contained  in  17,3 
times  and  2  over,  17  is  the  dividend,  5  the  divisor,  3  the 
quotient,  and  2  the  remainder. 

NOTES. — i.  The  term  division,  is  from  the  Latin  divido,  to  part, 
or  divide. 

The  term  dividend,  from  the  same  root,  signifies  that  which  is  to  be 
divided  ;  the  termination  nd,  having  the  force  of  to  be.  (Art.  33,  n.) 

Quotient  is  from  the  Latin  quoties,  signifying  how  often  or  how 
many  times. 

2.  The  remainder  is  always  the  same  denomination  as  the  divi- 
dend ;  for  it  is  an  undivided  part  of  it. 

3.  A  proper  remainder  is  always  less  than  the  divisor. 

61.  An  obvious  way  to  find  how  many  times  the  divisor 
is  contained  in  the  dividend,  is  to  subtract  the  divisor 
from  the  dividend  continually,  till  the  latter  is  exhausted, 
or  till  the  remainder  is  less  than  the  divisor ;  the  number 
of  subtractions  will  be  the  quotient.     Thus,  taking  any 
two  numbers,  as  4  and  12,  we  have  12—4=8;  8—4=4; 
and  4—4=0.    Here  are  three  subtractions;  therefore,  4  i8 
contained  in  12,  3  times.     Hence, 

Division  is  sometimes  said  to  be  a  sliort  method  of  con- 
tinued subtraction. 

60.  What  is  division  ?  The  number  to  be  divided  called  ?  To  divide  by  ?  Tho 
result?  Tlip  r>art  left  ?  Note.  Mr-anine-  of  the  term  division  1  Dividend?  Quo- 
tient ?  What  denomination  is  the  remainder  ?  Why  ? 


54 


DIVISION. 


But  there  is  a  shorter  and  more  direct  way  of  obtaining 
the  quotient.  For  we  know  by  the  multiplication  table, 
that  3  times  4,  or  4  taken  3  times,  are  1 2 ;  hence,  4  is 
contained  in  1 2,  3  times. 

62.  Division  is  the  reverse  of  multiplication.  In  mul- 
tiplication loth  factors  are  given,  and  it  is  required  to  find 
the  product ;  in  division,  one  factor  and  the  product  (which 
answers  to  the  dividend)  are  given,  and  it  is  required  to 
find  the  other  factor,  which  answers  to  the  quotient. 
Hence,  division  may  be  said  to  be  finding  a  quotient  which, 
multiplied  into  the  divisor,  will  produce  the  dividend. 

NOTE. — When  the  dividend  contains  only  one  denomination,  the 
operation  is  called  Simple  Division. 

DIVISION    TABLE. 


i  is  in 

2  is  in 

3  is  in 

4  is  in      5  is  in     6  is  in 

i,  once. 

2,  once. 

3,  once. 

4,  once.     5,  once.    6,  once. 

2,              2 

4,         2 

6,              2 

8,              2   10,              2  12,           2 

3,         3 

6,         3 

9,          3 

12,          3  15,          3  18,        3 

4,         4 

8,          4 

12,          4 

16,          4  20,          4  24,        4 

5,         5 

io,          5 

15,          5 

20,          5  25,          5  30,        5 

6,         6 

12,              6 

18,          6 

24,          630,          636,        6 

7,         1 

14,          7 

21,          7 

28,          7l35,          742,        7 

8,          8 

16,          8 

24,          8 

32,          840,          848,        8 

9,          9 

18,          9 

27,          9 

36,          9  45,          9  54,        9 

10,        10 

20,        10 

30,        10 

40,        1050,        1060,      10 

7  i*  in 

8  is  in 

9  is  in 

10  is  in 

ii  is  in 

12  is  in 

7,  once. 

8,  once. 

9,  once. 

10,  once. 

1  1,  once. 

12  once. 

14,         2 

16,          2 

18,        2 

20,           2 

22,           2 

24,       2 

21,         3 

24,         3 

27,        3 

3°,        3 

33,       3 

36,        3 

28,          4 

32,          4 

36,        4 

40,        4 

44,        4 

48,        4 

35,         5 

40,         5 

45,        5 

5°,       5 

55,       5 

60,        5 

42,         6 

48,          6 

54,        6 

60,        61  66,        6 

72,        6 

49,          7 

56,         7 

63,        7 

7°,        7 

77,  '      7 

84,        7 

56,          8 

64,          8 

72,        8 

80,        8 

88,        8 

96,        8 

63>          9 

72,          9 

81,        9 

90,        9 

99,        9 

1  08,        9 

70,        1080,        10 

90,      10 

IOO,         IO 

IIO,         10 

1  2O,        IO 

61.  What  is  an  obvious  way  to  find  how  many  times  the  divisor  IB  contained 
In  the  dividend  ?    What  is  division  sometimes  called  t 


DIVISION.  05 

OBJECTS    OF    DIVISION. 
63.  The  object  or  office  of  Division  is  twofold: 
ist,  To  find  how  many  times  one  number  is  contained  in 
another. 

zd,  To  divide  a  number  into  equal  parts. 

63,  <(•  To  find  how  many  times  one  number  is  contained 

in  another. 

i.  A  man  has  15  dollars  to  lay  out  in  books,  which  are 
3  dollars  apiece  :  how  many  can  he  buy  ? 

ANALYSIS.  —  In  this  problem  the  object  is  to  find  Jww  many  times 
3  dols.  are  contained  in  15  dols. 

Let  the  15  dols.  be  represented  by  15  counters,  or  unit  marks. 
Separating  these  into  groups  of  3  each,  there  are  5  groups.  There- 
fore, he  can  buy  5  books. 


63,  &•      To  divide  a  number  into  equal  parts. 

2.  If  a  man  divides  15  dollars  equally  among  3  persons, 
how  many  dollars  will  each  receive  ? 

ANALYSIS.  —  The  object  here  is  to  divide  15  dols.  into  3  equal  parts. 

Let  the  15  dols.  be  represented  by  15  counters.  If  we  form  3  groups, 
first  putting  I  counter  in  each,  then  another,  till  the  counters  are  ex- 
hausted, each  group  will  have  5  counters.  Therefore,  each  person 
will  receive  5  dollars. 

^oo^^lo^^^ol^^-^^^ 

REMARK.  —  The  preceding  are  representative  examples  of  the  two 
classes  of  problems  to  which  Division  is  applied.  In  the  first,  the 
divisor  and  dividend  are  the  same  denomination,  and  the  quotient  is 
times,  or  an  abstract  number. 

In  the  second,  the  divisor  and  dividend  are  different  denominations, 
and  the  quotient  is  the  same  denomination  as  the  dividend.  Hence, 

64.  When  the  divisor  and  dividend  are  the  same  denom- 
ination, the  quotient  is  always  an  abstract  number. 

62.  Of  what  is  division  the  reverse  ?  What  is  given  in  multiplication  ?  What 
required?  What  is  given  in  division?  What  required?  Note.  When  the  divi- 
dend contains  but  one  denomination,  what  is  the  operation  called  ?  63.  What  is 
the  ohject  or  office  of  division  ?  63,  a.  What  is  the  object  in  the  first  problem  f 
63,  b.  What  in  the  second  ?  Bern.  What  is  said  of  first  two  problems  I 


DIVISION. 

When  the  divisor  and  dividend  are  different  denomina- 
tions, the  quotient  is  always  the  same  denomination  as  the 
dividend. 

NOTES. — i.  The  process  of  separating  a  number  into  equal  parts,  as 
required  in  the  second  class  of  problems,  gave  rise  to  the  name 
"  Division."  It  is  also  the  origin  of  Fractions.  (Art.  134.) 

2.  The  mode  of  reasoning  in  the  solution  of  these  two  classes  of 
examples  is  somewhat  different ;  but  the  practical  operation  is  the 
same,  viz. :  to  find  7iow  many  times  one  number  is  contained  in 
another,  which  accords  with  the  definition  of  Division. 

65.  When  a  number  or  thing  is  divided  into  two  equal 
parts,  the  parts  are  called  halves;  into  three,  the  parts  are 
called  thirds  ;  into  four,  they  are  called  fourths  ;  etc. 

The  number  of  parts  is  indicated  by  their  name. 

66.  A  number  is  divided  into  two,  three,  four,  five,  etc., 
equal  parts  by  dividing  it  by  2,  3,  4,  5,  etc.,  respectively. 

3.  What  is  a  half  of  12  ?    A  third  of  15  ?    A  fourth  of 
20  ?    A  fifth  of  35  ?    A  seventh  of  28  ? 

4.  What  is  a  sixth  of  42  ?    A  seventh  of  56  ?    A  ninth 
of  63  ?    An  eighth  of  72  ?  "  A  twelfth  of  108  ? 

67.  The  Siffn  of  Division  is  a  short  horizontal 
line  between  two  dote   (-T-),  placed  before  the  divisor. 
Thus,  the  expression  28-7-7,  shows  that  28  is  to  be  divided 
by  7,  and  is  read,  "28  divided  by  7." 

68.  Division  is  also  denoted  by  writing  the  divisor 
under  the  dividend,  with   a  short   line  between  them. 
Thus,  -2,§-  is  equivalent  to  28-5-7.     I*  is  rea^>  "  2&  divided 
by  7,"  or  "28  sevenths." 

69.  Division  is  commonly  distinguished  as  Short  Divis~ 
ion  and  Long  Division. 

64.  When  the  divisor  and  dividend  are  the  same  denomination,  what  is  the 
quotient  ?  When  different  denominations,  what  ?  65.  When  a  number  is  divided 
into  two  equal  parts,  what  are  the  parts  called  ?  Into  three  ?  Four  ?  66.  How 
divide  a  number  into  2,  3,  4,  etc.,  equal  parts?  67.  What  is  the  sitrn  of  division? 
What  does  the  expression  35-1-7  show?  68.  How  else  is  division  denoted? 


SHOET    DIVISION. 

70.  Short  Division  is  the  method  of  dividing, 
when  the  results  of  the  several  steps  are  carried  in  the 
mind,  and  the  quotient  only  is  set  down. 

71.  To  divide  by  Sliort  Division. 

Ex.  i.  If  apples  are  $3  a  barrel,  how  many  barrels  can 
you  buy  for  $693  ? 

ANALYSIS. — Since  $3  will  buy  i  barrel,  $693         OPERATIC*. 
will  buy  as  many  barrels  as  $3  are  contained        §^$>6ot 
times  in  $693.    Let  the  numbers  be  set  down  as 
in  the  margin.    Beginning  at  the  left,  we  proceed       **uot-  23*  °Kt" 
thus :  3  is  contained  in  6  hundred,  2  hundred  times. 
Set  the  2  in  hundreds'  place,  under  the  figure  divided,  became  it  is 
hundreds.    Next,  3  is  contained  in  9  tens,  3  tens  times,     feet  the  3 
in  tens'  place  under  the  figure  divided.    Finally,  3  is  contained  in 
3  units,  i  time.    Set  the  i  in  units'  place.    Ans.  231  barrels. 

REM. — This  problem  belongs  to  the  ist  class,  the  object  being  to 
find  how  many  times  one  number  is  contained  in  another.  (Art.  63,  a.) 

Solve  the  following  examples  in  the  same  manner : 

(2.)  (3-)  (4-)  (5-) 

2)4468  3)3696  4)4848  5)5555 

6.  A  man  having  27543  pounds  of  grapes,  packed  them 
for  market  in  boxes  containing  5  pounds  each :  how  many 
boxes  did  he  fill,  and  how  many  pounds  over? 

ANALYSIS. — He  used  as  many  boxes  as  there          OPERATION. 
are  times  5  pounds  in  27543  pounds.     Let  the       5)27543  pounds, 
numbers  be  set  down  as  in  the  margin.     Since     ouot.  cco8~boxes, 
the  divisor  5  is  not  contained  in  the  first  figure  and    p  over 

of  the  dividend,  we  find  how  many  times  it  is 
contained  in  the  first  two  figures,  which  is  5  times  and  2  over.  We 
set  the  quotient  figure  5  under  the  right  hand  figure  divided, 
because  it  is  the  same  order  as  that  figure,  and  prefix  the  remainder 
2,  mentally  to  the  next  figure  of  the  dividend,  making  25.  Now  5 
is  in  25,  5  times,  and  no  remainder.  Again,  5  is  not  contained  in  4, 
the  next  figure  of  the  dividend  ;  we  therefore  place  a  cipher  in  the 


58  DIVISION. 

quotient,  and  prefix  the  4  mentally  to  tlie  next  figure  of  the  dividend, 
as  if  it  were  a  remainder,  making  43.  Finally,  5  is  in  43,  8  times, 
and  3  over.  He  therefore  filled  5508  boxes,  and  had  3  pounds 
remainder.  Hence,  the 

EULE.— I.  Place  the  divisor  on  the  left  of  the  dividend, 
and  beginning  at  the  left,  divide  each  figure  by  it,  setting 
the  result  under  the  figure  divided. 

II.  If  the  divisor  is  not  contained  in  a  figure  of  the  div- 
idend, put  a  cipher  in  the  quotient,  and  find  hoiv  many 
times  it  is  contained  in  this  and  the  next  figure,  setting  the 
result  under  the  right  hand  figure  divided. 

III.  If  a  remainder  arise  from  any  figure  before  the  last, 
prefix  it  mentally  to  the  next  figure,  and  divide  as  before. 

If  from  the  last  figure,  place  it  over  the  divisor,  and 
annex  it  to  the  quotient. 

NOTES. — i.  In  the  operation,  the  divisor  is  placed  on  the  left  of 
the  dividend,  and  the  quotient  under  it,  as  a  matter  of  convenience. 
When  division  is  simply  represented,  the  divisor  is  either 
placed  under  the  dividend,  or  on  the  right,  with  the  sign  (-J-)  be- 
fore it. 

2  The  reason  for  beginning  to  divide  at  the  left  hand  is,  that  in 
dividing  a  higher  order  there  may  be  a  remainder,  which  must  be 
prefixed  to  the  next  lower  order,  as  we  proceed  in  the  operation. 

3.  We  place  the  quotient  figure  under  the  figure  divided;  because 
the  former  is  the  same  order  as  the  latter.    (Art.  71.) 

4.  When  the  divisor  is  not  contained  in  a  figure  of  the  dividend, 
we  place  a  cipher  in  the  quotient,  to  show  that  the  quotient  has  no 
units  corresponding  with  the  order  of  this  figure.    It  also  preserves 
the  local  value  of  the  subsequent  figures  of  the  quotient. 

5.  The  final  remainder  shows  that  a  part  of  the  dividend  is  not 
divided.     It  is  placed  over  the  divisor  and  annexed  to  the  quotient 
to  complete  the  division. 

70.  What  is  Short  Division?  71  How  write  nnmters  for  short  division? 
The  next  step  ?  When  the  divisor  is  not  contained  in  a  figure  of  the  dividend, 
how  proceed  ?  When  there  is  a  remainder  after  dividing  a  figure,  how  ?  If  there  is 
a  remainder  after  dividing  the  last  figure,  what  ?  Note.  Why  place  the  divisor  on 
the  left  of  the  dividend  ?  Why  begin  to  divide  at  the  left  hand  ?  Why  place  each 
quotient  figure  under  the  figure  divided?  Why  place  a  cipher  in  the  quotient, 
when  tho  divisor  is  not  contained  tn  a  figure  of  the  dividend  ?  Whnt  doos  the  final 
remainder  show  ?  Why  place  it  over  the  divisor  and  annex  it  to  the  Quotient? 


DIVISION.  59 

53^"  The  pupil  should  early  learn  to  abbreviate  the  language  used 
In  the  process  of  dividing.  Thus,  in  the  next  example,  instead  of 
eaying  5  is  contained  in  7  once,  and  2  over,  let  him  pronounce  the 
quotient  figures  only;  as,  one,  four,  five,  seven. 

i.  A  man  divided  7285  acres  of  land  equally  among  his 
5  sons  :  what  part,  and  how  much,  did  each  receive  ? 

SOLUTION.  —  i  is  i  fifth  of  5  ;  hence  each  had  i  fifth  part.  (Art.  65.) 
Again,  7285  A  -5-5  =  1457  A;  hence  each  received  1457  A.  (Art.  66.) 


(2.) 

(3-)                       (40                      (50 
3)560346              4)689034             5)748239 

(6.) 

6)3972647 

(7-)                      (8.)                       (9.) 
7)4806108           8)739°464           9)8306729 

(10.) 

10)57623140 

(II.)                                                 (12.) 
11)66730145!                        12)8160252397 

13.  At  $2  apiece,  how  many  hats  can  be  bought  for 
$16486  ?    (Art  48,  Note  3.) 

14.  At  $4  a  head,  how  many  sheep  can  be  bought  for 
$844? 

15.  How  many  times  are  3  rods  contained  in  26936 
rods? 

1  6.  A  man  having  $42684,  divided  it  equally  among 
his  4  children  :  how  much  did  each  receive  ? 

17.  If  a  quantity  of  muslin  containing  366  yards  is  di- 
vided into  3  equal  parts,  how  many  yards  will  each  part 
contain  ? 

1  8.  If  84844  pounds  of  bread  are  divided  equally  among 
6  regiments,  how  many  pounds  will  each  regiment  re- 
ceive ? 

19.  Eight  men  found  a  purse  containing  $64968,  which 
they  shared  equally:  how  much  did  each  receive? 


60  DIVISION. 

20.  Divide  4268410  by  4.  21.  Divide  5601234  by  6. 

22.  Divide  6403021  by  5.  23.  Divide  7008134  by  7. 

24.  Divide  8210042  by  n.  25.  Divide  9603048  by  8. 

26.  Divide  23468420  by  10.  27.  Divide  32064258  by  9. 

28.  Divide  46785142  by  8.  29.  Divide  59130628  by  7. 

30.  Divide  653000638  by  n.  31.  Divide  774230029  by  12 

32.  In  7  days  there  is  i  week:    how  many  weeks  in 
26563  days? 

33.  If  $38472  are  divided  equally  among  6  persons,  how 
much  will  each  receive  ? 

34.  How  many  tons  of  coal,  at  $7  a  ton,  can  be  pur- 
chased for  ,$63456  ? 

35.  At  $9  a  barrel,  how  much  flour  can  be  bought  for 
$47239? 

36.  In  12  months  there  is  i  year:  how  many  years  in 
41260  months? 

37.  A  merchant  laid  out  $45285  in  cloths,  at  $7  a  yard: 
how  many  yards  did  he  buy  ? 

38.  At  8  shillings  to  a  dollar,  how  many  dollars  are 
there  in  75240  shillings? 

39.  In  9  square  feet  there  is  i  square  yard :  how  many 
square  yards  are  there  in  52308  square  feet? 

40.  If  a  person  travel  10  miles  an  hour,  how  long  will 
it  take  him  to  travel  25000  miles  ? 

41.  A  market  woman  having  845280  eggs,  wished  to 
pack  them  in  baskets  holding  i  dozen  each :  how  many 
baskets  did  it  take  ? 

42.  If  a  prize  of  $116248  is  divided  equally  among  8 
men,  what  will  be  each  one's  share  ? 

43.  If  in  7  townships  there  are  2346281  acres,  how  many 
acres  are  there  to  a  township  ? 

44.  At  $8  a  barrel,  how  many  barrels  of  sugar  can  be 
bought  for  $i  1 1364  ? 

45.  At  $11  each,  how  many  cows  can  be  had  for  $88990  ? 


"  ' 


LOKG   Division*. 

72.  Long  Division  is  the  method  of  dividing,  when 
the  results  of  the  several  steps  and  the  quotient  are  both 
set  down. 

73.  To  divide  by  Long  Division. 

i.  A  speculator  paid  $31097  for  15  city  lots:  what  did 
the  lots  cost  apiece  ? 

ANALYSIS.—  Since  15  lots  cost  $31097,  I  OPERATION. 

lot  will  cost  as  many  dollars  as  15  is  con- 
tained  times  in  31097.    Let  the  divisor  and  3° 

dividend  be  set  down  as  in  the  margin.  — 

Beginning  at  the  left,  the  first  step  is  to  109 

find  how  many  times  the  divisor   15,   is  IQC 

contained  in  31  (which  is  2  times),  and  set  _ 

the  2  on  the  right  of  the  dividend.  x  7 

Second,  multiply  the    divisor   by   the  ** 

quotient  figure,  and  set  the  product  30,  _ 

under  the  figures  divided.     TJiird,  sub-  2 

tract    this     product    from    the     figures 

divided.  FourtJi,  bring  down  the  next  figure  of  the  dividend, 
and  place  it  on  the  right  of  the  remainder,  making  10  for  tho 
next  partial  dividend,  and  proceed  as  before.  But  15  is  not  con- 
tained in  10  ;  we  therefore  place  a  cipher  in  the  quotient,  and 
bring  down  the  next  figure  of  the  dividend,  making  109.  Now 
15  is  in  109,  7  times.  Set  the  quotient  figure  7  on  the  right,  mul- 
tiply the  divisor  by  it,  subtract  the  product  from  the  partial  divi- 
dend, and  to  the  right  of  the  remainder,  bring  down  the  succeeding 
figure  for  the  next  partial  dividend,  precisely  as  before.  Now  15  is 
in  47,  3  times.  Setting  the  3  in  the  quotient,  multiplying,  and  sub- 
tracting, as  above,  the  final  remainder  is  2.  We  place  this  remainder 

72.  What  is  Long  Division?  73.  How  write  the  numbers?  What  is  the  first 
Btep?  The  second?  The  third?  The  fourth?  If  the  divisor  is  not  contained 
in  a  partial  dividend,  how  proceed?  What  is  to  be  done  with  the  last  remainder? 
Note.  What  is  the  difference  between  short  and  long  division  ?  Of  what  order  itt 
the  quotient  figure  ? 


62  DIVISION. 

over  the  divisor  and  annex  it  to  the  quotient.  The  divisor  and  divi- 
dend being  different  denominations,  the  quotient  is  the  same  as  the 
dividend  (Art.  64).  Therefore  the  lots  cost  $2073  -&  apiece.  Hence,  the 

EULE.— I.  Find  how  many  times  the  divisor  is  contained 
in  the  fewest  figures  on  the  left  of  the  dividend,  that  will 
contain  it,  and  set  the  quotient  on  the  right. 

II.  Multiply  the  divisor  by  this  quotient  figure,  and  sub- 
tract the  product  from  the  figures  divided. 

III.  To  the  right  of  the  remainder,  bring  down  the  next 
figure  of  the  dividend,  and  divide  as  before. 

IV.  If  the  divisor  is  not  contained  in  a  partial  dividend, 
place  a  cipher  in  the  quotient,  bring  down  another  figure, 
and  continue  the  operation  till  all  the  figures  are  divided. 

If  there  is  a  remainder  after  dividing  the  last  figure,  set 
it  over  the  divisor,  and  annex  it  to  the  quotient. 

NOTES. — i.  The  parts  into  which  the  dividend  is  separated  in 
finding  the  quotient  figure,  are  called  partial  dividends,  because 
they  are  parts  of  the  whole  dividend. 

2.  Short  and  Long  Division,  it  will  be  seen,  are  the  same  in  prin- 
ciple.    The  only  difference  is,  that  in  one  the  results  of  the  several 
steps  are  carried  in  the  mind,  in  the  other  they  are  set  down. 

Short  Division  is  the  more  expeditious,  and  should  be  employed 
when  the  divisor  does  not  exceed  12. 

The  reasons  for  the  arrangement  of  the  parts,  and  for  beginning 
to  divide  at  the  left  hand,  are  the  same  as  in  Short  Division. 

3.  The  quotient  figure  in  Long  as  well  as  in  Short  Division,  is 
always  of  the  same  order  as  that  of  the  right  hand  figure  of  the  par- 
tial dividend. 

4.  To  prevent  mistakes,  it  is  customary  to  place  a  mark  under 
the  several  figures  of  the  dividend  as  they  are  brought  down. 

5.  After  the  first  quotient  figure  is  obtained,  for  each  succeeding 
figure  of  the  dividend,  either  a  significant  figure  or  a  cipher  must  be 
put  in  the  quotient. 

6.  If  the  product  of  the  divisor  into  the  figure  placed  in  the  quo- 
tient is  greater  than  the  partial  dividend,  it  is  plain  the  quotient 
figure  is  too  large,  and  therefore  must  be  diminished. 

If  the  remainder  is  equal  to  or  greater  than  the  divisor,  the  quo- 
tient figure  is  too  small,  and  must  be  incrcawL 


DIVISION.  63 


PROOF. 

74.  By  Multiplication. — Multiply  the  divisor  and  quo- 
tient together,  and  to  the  product  add  the  remainder.    If 
the  result  is  equal  to  the  dividend,  the  work  is  right. 

NOTE. — This  proof  depends  upon  the  principle,  that  Division 
is  the  reverse  of  Multiplication  ;  the  dividend  answering  to  the  pro- 
duct, the  divisor  to  one  of  the  factors,  and  the  quotient  to  the  other. 
(Art.  62.) 

75.  By   excess   of  98. — Multiply  the  excess  of  gs  in 
the  divisor  by  that  in  the  quotient,  and  to  the  product  add 
the  remainder.     If  the  excess  of  gs  in  this  sum  is  equal  to 
that  in  the  dividend,  the  ivork  is  right. 

2.  Divide  181403  by  67,  and  prove  the  operation. 

Ans.  270714. 

Proof. — By  Multiplication. — 2707x67=181369,  and  181369  +  34 
the  rem.  =  181403  the  dividend. 

By  excess  of  gs. — The  excess  of  gs  in  the  divisor  is  4,  and  the 
excess  in  the  quotient  is  7.  Now  4x7=28,  and  28  +  34=62;  the 
excess  of  gs  in  62  is  8.  The  excess  of  93  in  the  dividend  is  also  8. 

3.  Divide  34685  by  15.          4.  Divide  65456  by  16. 

5.  Divide  41534  by  20.          6.  Divide  52663  by  25. 

7.  Divide  420345  by  39.        8.  Divide  506394  by  47. 

9.  Divide  673406  by  69.       10.  Divide  789408  by  77. 
ii.  Divide  4375023  by  86.     12.  Divide  5700429  by  93. 
13.  Divide  6004531  by  59.     14.  Divide  8430905  by  78. 
15.  Divide  7895432  by  89.     16.  Divide  9307108  by  98. 

17.  How  many  acres  of  land  at  $75  per  acre,  can  I  buy 
for  $18246  ? 

1 8.  At  $83  apiece,  how  many  ambulances  can  be  bought 
for  $37682  ? 

73.  Note.  If  the  product  of  the  divisor  into  the  figure  placed  in  the  quotient,  is 
greater  than  the  partial  dividend,  what  does  it  show  ?  If  the  remainder  is  equal 
to  or  greater  than  the  divisor,  what  f  How  is  Division  proved  ? 


64  DIVISION. 


76.  To  find  the  Quotient  Figure,  when  the  Divisor  is  large. 

Take  t\\e  first  figure  of  the  divisor  for  a  trial  divisor,  and 
find  how  many  times  it  is  contained  in  the  first  or  first 
two  figures  of  the  dividend,  making  due  allowance  for 
carrying  the  tens  of  the  product  of  the  second  figure  of  the 
divisor  into  the  quotient  figure. 

19.  Divide  18046  by  673. 

ANALYSIS. — Taking  6  for  a  trial  divisor,  it  is     673)18046(26 

contained  in  18,  3  times.    But  in  multiplying  7  by  1346 
3,  there  are  2  to  carry,  and  2  added  to  3  times  6, 

make  20.     But  20  is  larger  than  the  partial  divi  4586 

dend  18  ;  therefore,  3  is  too  large  for  the  quotient  4038 
figure.     Hence,  we  place  2  in  the  quotient,  and 

proceed  as  before.     (Art.  73,  n.)  c^g 

20.  Divide  3784123  by  127.     21.  Divide  436 17 29  by  219. 
22.  Divide  8953046  by  378.     23.  Divide  9073219  by  738. 

24.  How  many  shawls  at  $95,  can  be  bought  for  $42750  ? 

25.  In  144  square  inches  there  is  i  square  foot:  how 
many  square  feet  are  there  in  59264  square  inches  ? 

26.  A  quartermaster  paid  $29328  for  312  cavalry  horses: 
how  much  was  that  apiece  ? 

27.  If  128  cubic  feet  of  wood  make  i  cord,  how  many 
cords  are  in  69240  cubic  feet? 

28.  If  a  purse  of  $150648  is  divided  equally  among  250 
sailors,  how  much  will  each  receive  ? 

29.  In  1728  cubic  inches   there  is  i  cubic  foot:   how 
many  cubic  feet  are  there  in  250342  cubic  inches  ? 

30.  If  560245   pounds  of  bread  are  divided  equally 
among  11200  soldiers,  how  much  will  each  receive? 

31.  Div.  36942536  by  4204.      32.  Div.  57300652  by  5129. 
33.  Div.  629348206  by  52312.    34.  Div.  730500429  by  61073, 

35.  Divide  7300400029  by  236421. 

36.  Divide  8230124037  by  463205. 

37.  Divide  843000329058  by  203963428. 


DIVISION.  65 

38.  A  stock  company  having  $5000000,  was  divided 
into  1250  shares:  what  was  the  value  of  each  share? 

39.  A  railroad  478  miles  in  length  cost  $18120000 :  what 
was  the  cost  per  mile  ? 

40.  A  company  of  942  men  purchased  a  tract  of  land 
containing  272090  acres,  which  they  shared  equally:  what 
was  each  man's  share  ? 

41.  A  tax  of  $42368200  was  assessed  equally  upon  5263 
towns :  what  sum  did  each  town  pay  ? 

42.  The  government  distributed  $9900000  bounty  equally 
among  36000  volunteers:  how  much  did  each  receive? 

43.  Since  there  is  i  year  in  525600  minutes,  how  many 
years  are  there  in  105192000  minutes. 

CONTRACTIONS. 
CASE   I. 

77.  To  divide  by  a  Composite  Number. 

Ex.  i.  A  farmer  having  300  pounds  of  butter,  packed  it 
in  boxes  of  15  pounds  each :  it  is  required  to  find  how 
many  boxes  he  had,  using  the  factors  of  the  divisor. 

ANALYSIS. — 15=5  times  3.  Now  if  he  puts 
5  pounds  into  a  box,  it  is  plain  he  would  use 
as  many  boxes  as  there  are  53  in  300  or  60  OI 

fine-pound  boxes.     But   3  five-pound  boxes          5/3°°  lba- 
make   i  fifteen-pound  box ;    hence,  60  fife-  3)60  5  lb.  boxes. 

pound  boxes  will  make  as  many  15  pound    _Ans.    20  boxes. 
boxes,  as  3  is  contained  times  in  60.  which  is 
20.    In  the  operation,  we  first  divide  by  one 
of  the  factors  of  15,  and  the  quotient  by  the  other.    Hence,  the 

EULE. — Divide  the  dividend  ~by  one  of  the  factors  of  the 
divisor,  and  the  quotient  thence  arising  by  another  factor, 
and  so  on,  until  all  the  factors  have  been  used.  TJie  last 
quotient  will  be  the  one  required. 

77.  When  the  divisor  Is  a  composite  number,  how  procoed  T 


66  DIVISION. 

NOTES. — I.  This  contraction  is  the  reverse  of  multiplying  by  a 
composite  number.  Hence,  dividing  the  dividend  (which  answers  to 
the  product)  by  the  several  factors  of  one  of  the  numbers  which  pro- 
duced it,  will  evidently  give  the  quotient,  the  other  factor  of  which 
the  dividend  is  composed.  (Art.  56.) 

2.  When  the  divisor  can  be  resolved  into  different  sets  of  factors, 
the  result  will  be  the  same,  whichever  set  is  taken,  and  whatever 
the  order  in  which  they  are  employed.  The  pupil  is  therefore  at 
liberty  to  select  the  set  and  the  order  most  convenient. 

2.  Divide  357  by  21,  using  the  factors. 

3.  If  532  oranges  are  divided  equally  among  28  boys, 
what  part,  and  how  many  will  each  receive  ? 

4.  A  dairyman  packed  805  pounds  of  butter  in  35  jars : 
how  many  pounds  did  he  put  in  a  jar  ? 

5.  How  many  companies  in  a  regiment  containing  756 
soldiers,  allowing  63  soldiers  to  a  company  ? 

6.  Divide  204  by  12,  using  different  sets  of  factors. 

7.  Divide  368  by  16,  using  different  sets  of  factors. 

8.  Divide  780  by  30,  using  different  sets  of  factors. 

78.  To  find  the   True  Remainder,  when   Factors  of  the 
divisor  are  used. 

Ex.  9. — A  lad  picked  2425  pints  of  chestnuts,  which  he 
wished  to  put  into  bags  containing  64  pints  each:  it  is 
required  to  find  the  number  of  bags  he  could  fill,  and  the 
number  of  pints  over,  or  the  true  remainder. 

ANALYSIS.  —  The   divisor   64 
equals  the  factors  2x8x4.    Di- 

.,.  .     ,       ,  ,,  OPERATION. 

vidmg  2425  pints  by  2,  the  quo-  2^2A2C 

tient  is   1212  and  i  remainder.  J- — 

But  the  units  of  the  quotient  1212  8)I2I2~  *  J  P*- I8t r- 

are   2   times  as    large  as  pints,  4)151 — 45  4x2=        8pt.  2dr. 

which,  for  the  sake  of  distinction,          -77 ^  •  i  x  8  X  2  —  48  pt.  *d  r 

we  will  call  quarts;  and  the  re- 
mainder i,  is  a  pint,  the  same  as  An^  ^  b       and  s ;   t 
the  dividend.     (Art.  60,  n.) 

Next,  dividing  1212  quarts  by 
8,  the  quotient  is  151,  and  4  remainder.    But  the  units  of  the 

78.  How  is  the  true  remainder  fonnd  1  Note.  To  what  is  It  eqnnl  ?  \Vh»t 
should  he  done  with  it?  The  oh.iect  In  multiplying  each  partial  remainder  by  all 
the  preceding  divisors  except  its  own  ? 


DIVISION.  67 

quotient  151  are  8  times  as  large  as  quarts;  call  them  pecks ;  the 
remainder  4,  which  denotes  quarts,  must  be  multiplied  by  the 
preceding  divisor  2,  to  reduce  it  back  to  pints.  Finally,  dividing  151 
pecks  by  4,  the  quotient  is  37,  and  3  remainder.  But  the  units  of 
the  quotient  37  have  four  times  the  value  of  pecks  ;  call  them  bags  ; 
and  the  remainder  3,  which  denotes  pecks,  must  be  multiplied  by 
the  last  divisor  8,  to  reduce  them  back  to  quarts,  and  be  multiplied 
by  2  to  reduce  the  quarts  to  pints.  Having  filled  37  bags,  we  have 
three  partial  remainders,  I  pint,  4  quarts,  and  3  pecks. 

The  next  step  is  to  find  the  true  remainder.  We  have  seen  that 
a  unit  of  the  2d  remainder  is  twice  as  large  as  those  of  the  given 
dividend,  which  are  pints  ;  hence,  4  quarts  =  8  pints.  Again,  each 
unit  of  tlve  3d  remainder  is  8  times  as  large  as  those  of  the  preceding 
dividend,  which  are  quarts  ,  therefore,  3  pecks  =  3  x  8  or  24  quarts  ; 
and  24  qU.—  24  x  2  or  48  pts.  The  sum  of  these  partial  remainders, 
i  pt.  +  8  pt,  +  48  pt.  =  57  pts.,  is  the  true  remainder.  Hence,  the 

KULE. — Multiply  each  partial  remainder  by  all  the 
divisors  preceding  its  own  ;  the  sum  of  these  results  added 
to  the  first,  will  be  the  true  remainder. 

NOTES. — i.  Multiplying  each  remainder  by  all  the  preceding  di- 
visors except  its  own,  reduces  them  to  units  of  the  same  denomina- 
tion as  the  given  dividend.  Hence,  the  true  remainder  is  equal  to 
the  sum  of  the  partial  remainders  reduced  to  the  same  denomina- 
tion as  the  dividend. 

2.  When  found,  it  should  be  placed  over  the  given  divisor,  and  be 
annexed  to  the  quotient. 

10.  Divide  43271  by  45.          n.  Divide  502378  by  63. 
12.  Divide  710302  by  72.        13.  Divide  3005263  by  84. 
14.  Divide  63400511  by  96.    15.  Divide  216300265  by  144. 

CASE   II. 

79.  To  divide  by  10,  100,  1000,  etc. 

Ex.  1 6.  Divide  2615  by  100. 

ANALYSIS. — Annexing  a  cipher  to  a  figure,  we 
have  seen,  multiplies  it  by  10 ;   conversely,  re-  OTKHATIOB. 

moving  a  cipher  from  the  right  of  a  number     i|Qo)26  15 
must   diminish  its  value  10  times,  or  divide  it    Ans.  26  15  Rem. 
by  10;  for,  each  figure  in  the  number  is  re- 
moved one  plas.e  to  the  right.    (Art.  12.)    In  like  manner,  cutting 


68  DIVISION. 

off  two  figures  from  the  right,  divides  it  by  100 ;  cutting  off  three, 
divides  it  by  1000,  etc. 

In  the  operation,  as  the  divisor  is  100,  we  simply  cut  off  two 
figures  on  the  right  of  the  dividend ;  the  number  left,  viz.,  26,  is  the 
quotient;  and  the  15  cut  off,  the  remainder.  Hence,  the 

EULE. — From  the  right  of  the  dividend,  cut  off  as  many 
figures  as  there  are  ciphers  in  the  divisor.  The  figures  left 
tvill  be  the  quotient ;  those  cut  off,  the  remainder. 

17.  Divide  75236  by  100.  20.  9820341  by  100000. 

18.  Divide  245065  by  1000.        21.  9526401  by  1000000. 
79.  Divide  805211  by  10000.      22.  80043264  by  10000000. 

CASE    III. 

80.  To  divide, when  the  Divisor  has  Ciphers  on  the  right. 

23.  At  $30  a  barrel,  how  many  barrels  of  beef  can  be 
bought  for  $4273? 

ANALYSIS. — The  divisor  30  is  composed  of 
the  factors  3  and  10.     Hence,  in  the  operation, 
we  first  divide  by  10,  by  cutting  off  the  right-         3,°|4__7 [3 
hand  figure  of  the  dividend  ;  then  dividing  the     Quot.    142,  I  Eem. 
remaining  figures  of  the  dividend  by  3,  the      Ans.  I42vo  tls- 
quotient  is  142,  and  i  remainder.     Prefixing 
the  I  remainder  to  the  3  which  was  cut  off,  we  have  13  for  the  true 
remainder,  which  being  placed  over  the  given  divisor  30,  and  an- 
nexed to  the  quotient,  gives  142^  bis.  for  the  answer.     Hence,  the 

EULE. — I.  Cut  off  the  ciphers  on  the  right  of  the  divisor 
and  as  many  figures  on  the  right  of  the  dividend. 

II.  Divide  the  remaining  part  of  the  dividend  ly  the 
remaining  part  of  the  divisor  for  the  quotient. 

III.  Annex  the  figures  cut  off  to  the  remainder,  and  the 
result  will  be  the  true  remainder.     (Art.  78.) 

NOTE.— This  contraction  is  based  upon  the  last  two  Cases.  The 
1rue  remainder  should  be  placed  over  the  whole  divisor,  and  be  an- 
nexed to  the  quotient. 

79.  How  proceed  when  the  divisor  is  10,  100,  etc.  ?  80.  How  when  there  art 
ciphers  on  the  right  of  the  divisor?  Note.  Upon  what  is  this  contraction  based' 


DIVISION.  69 

«4.  Divide  45678  by  20.  25.  81386  by  200. 

26.  Divide  603245  by  3400.  27.  740321  by  6500. 

28.  Divide  7341264  by  87000.        29.  8004367  by  93000. 
30.  Divide  61273203  by  125000.     31.  416043271  by  670000. 

32.  In  100  cents  there  is  i  dollar:  how  many  dollars  in 
37300  cents  ? 

33.  At  $200  apiece,  how  many  horses  will  $45800  buy? 

34.  If  $75360  were  equally  distributed  among  1000  men, 

how  much  would  each  receive  ? 

35.  At  $4800  a  lot,  how  many  can  be  had  for  $25200? 

36.  How  many  bales,  weighing  450  pounds  each,  can  be 
made  of  27000  pounds  of  cotton  ? 


QUESTIONS    FOR    REVIEW,    INVOLVING    THE 
PRECEDING    RULES. 

1.  William,  who  has  219  marbles,  has  73  more  than 
James:  how  many  has  James  ?    How  many  have  both  ? 

2.  A  farmer  having  368  sheep,  wishes  to  increase  his 
flock  to  775  :  how  many  must  he  buy? 

3.  The  difference  of  two  persons'  ages  is  19  years,  and 
the  younger  is  57  years:  what  is  the  age  of  the  elder? 

4.  What  number  must  be  added  to  1368  to  make  3147  ? 

5.  What  number  subtracted  from  4118  leaves  1025  ? 

6.  What  number  multiplied  by  95  will  produce  7905  ? 

7.  The  product  of  the  length  into  the  breadth  of  a  field 
is  2967  rods, and  the  length  is  69  rods:  what  is  the  breadth? 

8.  A  man  having  5263  bushels  of  grain,  sold  all  but 
145  bushels:  how  much  did  he  sell? 

9.  What   number   must  be  divided  by   87,  that  the 
quotient  may  be  99  ? 

10.  If  the  quotient  is  217,  and  the  dividend  7595,  what 
must  be  the  divisor  ? 

11.  If  the  divisor  is  341  and  the  quotient  589,  what 
must  be  the  dividend  ? 


70  DIVISION. 

12.  A  merchant  bought  516  barrels  of  flour  at  $9  a  bar- 
rel, and  sold  it  for  $5275  :  how  much  did  he  gain  or  lose  ? 

13.  How  long  can  250  men  subsist  on  a  quantity  of 
food  sufficient  to  last  i  man  7550  days? 

14.  How  many  pounds  of  sugar,  at  n  cents,  must  be 
given  for  629  pounds  of  coffee,  at  17  cents? 

15.  At  $13  a  barrel,  how  many  barrels  of  flour  must  be 
given  for  530  barrels  of  potatoes  worth  $3  a  barrel  ? 

1 6.  A  man  having  $15260,  deducted  $4500  for  personal 
use,  and  divided  the  balance  equally  among  his  7  sons : 
how  much  did  each  son  receive  ? 

17.  A  man  earns  12  shillings  a  day,  and  his  son  8  shil- 
lings: how  long  will  it  take  both  to  earn  1200  shillings? 

1 8.  If  a  man  earns  $19  a  week,  and  pays  $2  a  week  fo* 
boarding  each  of  his  3  sons  at  school,  how  much  will  he 
lay  up  in  12  weeks  ? 

19.  A  farmer  sold  6  cows  at  $23,  150  bushels  of  wheat 
at  $2,  and  75  barrels  of  apples  at  $4,  and  laid  out  his  money 
in  cloth  at  $7  a  yard:  how  many  yards  did  he  have? 

20.  If  I  buy  1361  barrels  of  flour  at  $7,  and  sell  the 
whole  for  $12249,  how  much  shall  I  make  per  barrel  ? 

21.  The  earnings  of  a  man  and  his  two  sons  amount  to 
$3560  a  year;  their  expenses  are  $754.     If  the  balance  is 
divided  equally,  what  will  each  have  ? 

22.  A  man  having  $23268,  owed  $1733,  and  divided  the 
rest  among  four  charities :  how  much  did  each  receive  ? 

23.  How  many  sheep  at  $5  a  head,  must  be  given  for  30 
cows  at  $42  apiece  ? 

24.  A  father  bought  a  suit  of  clothes  for  each  of  his 
3  sons,  at  $123  a  suit,  and  agreed  to  pay  17  tons  of  hay  at 
$12  a  ton,  and  the  rest  in  potatoes  at  $43  barrel:  how 
many  barrels  of  potatoes  did  it  take  ? 

25.  If  you  add  $435  to  $567,  divide  the  sum  by  $334, 
multiply  the  quotient  by  217,  and  divide  the  product  by 
59,  what  will  be  the  result? 


DIVISION.  .  71 

26.  If  from  1530  you  take  319,  add  793  to  the  remainder, 
multiply  the  sum  by  44,  and  divide  the  product  by  37, 
what  will  be  the  result  ? 

27.  A  man's  annual  income  is  $4250 ;  if  he  spends  $1365 
for  house  rent,  $1439  for  other  expenses,  and  the  balance 
in  books,  at  $3  apiece,  how  many  books  can  he  buy  ? 

28.  A  farmer  having  $3038,  bought  15  tons  of  hay  at 
81 1, 3  yoke  of  oxen  at  $155, 375  sheep  at  $5,  and  spent  the 
rest  for  cows  at  $41  a  head :  how  many  cows  did  he  buy  ? 

GENERAL    PRINCIPLES    OF    DIVISION. 

81.  From  the  nature  of  Division,  the  absolute  value  of  the 
quotient  depends  both  upon  the  divisor  and  the  dividend. 

The  relative  value  of  the  quotient ;  that  is,  its  value  com- 
pared with  the  dividend,  depends  upon  the  divisor.  Thus, 

1.  If  the  divisor  is  equal  to  the  dividend,  the  quotient  is  i. 

2.  If  the  divisor  is   greater  than  the  dividend,  the 
quotient  is  less  than  i. 

3.  If  the  divisor  is  less  than  the  dividend,  the  quotient 
is  greater  than  i.    * 

4.  If  the  divisor  is  i,  the  quotient  is  equal  to  the  dividend. 

5.  If  the  divisor  is  greater  than  i,  the  quotient  is  less 
than  the  dividend. 

6.  If  the  divisor  is  less  than  i,  the  quotient  is  greater 
than  the  dividend. 

82.  The  relation  of  the  divisor,  dividend,  and  quotient 
is  such  that  the  divisor  remaining  the  same, 

Multiplying  the  dividend  by  any  number,  multiplies  the 
quotient  by  that  number.  Thus,  12-7-2=6;  and  (12x2) 
-7-2  =  6x2.  Conversely, 


81.  Upon  what  does  the  absolute  value  of  the  quotient  dapead?  Its  relative 
value  1  If  the  divisor  is  eqnal  to  the  dividend,  what  is  true  ok  the  quotient  ?  If 
greater?  If  less?  If  the  divisor  is  i,  what  is  the  quotient?  If  preater  tl.nn  i  T 
If  less  than  1 1  82.  Wliat  is  the  effect  of  multiplying  the  dividend  ?  83.  Of  divid- 
ing it? 


Tt  .  DIVISION. 

83.  Dividing  the  dividend  by  any  number,  divides  the 
quotient  by  that  number.     Thus,  (12-^- 2)-+ 2  =  6+ 2. 

84.  Multiplying  the  divisor  by  any  number,  divides  the 
quotient  by  that  number.     Thus,  24-7-6=4;    and  24-;- 
(6  x  2) =4-7-  2.     Conversely, 

85.  Dividing  the  divisor  by  any  number,  multiplies  the 
quotient  by  that  number.    Thus,  24-=- (6 4-2) =4  x  2. 

86.  Mulfyilying  or  dividing  both  the  divisor  and  rfwn- 
dfewc?  by  the  same  number,  does  not  «^er  the  quotient 
Thus,  48-7-8-6;  so  (48x2)-7-(8x2)  =  6;  and  (48^-2)^- 

(8-r-2)  =  6. 

NOTE. — Multiplying  or  dividing  the  dividend,  produces  a  like 
effect  on  the  quotient ;  but  multiplying  or  dividing  the  divisor,  pro- 
duces the  opposite  effect  on  the  quotient. 

86,  «•'  If  a  number  is  both  multiplied  and  divided  by  the 
same  number,  its  value  is  not  altered.  Thus,  (7  x  6)  ~-  6  is 
equal  to  7. 


PROBLEMS    AND    FORMULAS    PERTAINING    TO    THE 
FUNDAMENTAL    RULES. 

87.  A  Problem  is  something  to  be  done,  or  a  question 
to  be  solved. 

88.  A  Formula,  is  a  specific  rule  by  which  problems 
are  solved,  and  may  be  expressed  by  common  language,  or 
by  ,si^r«5. 

89.  The/owr  #raz£  Problems  of  Arithmetic  have  already 
been  illustrated.     They  are — 

ist.  When  two  or  more  numbers  are  given,  to  find  their 
sum,  or  amount.  (Art  29.) 

2d.  When  two  numbers  are  given,  to  find  their  difference. 

84.  What  is  the  effect  of  multiplying  the  divisor?  Of  dividing  it?  86.  WTiat 
is  the  effect  of  multiplying  or  dividing  both  the  divisor  and  dividend  ?  87.  What 
ia  a  problem?  S3.  A  formula?  89.  The  four  great  problems  ot  arithmetic!' 


PROBLEMS    AND    FORMULAS.  73 

3d.  When  two  factors  are  given,  to  find  their  product. 
(Art.  50.) 

4tb.  When  two  numbers  are  given,  to  find  how  many 
times  one  is  contained  in  the  other.  (Art.  73.) 

90.  These  problems  constitute  the  four  fundamental 
rules  of  Arithmetic,  called  Addition,  Subtraction,  Hulti~ 
plication,  and  Division. 

NOTES. — i.  These  rules  are  called  fundamental,  because  upon 
them  are  based  all  arithmetical  operations. 

2.  As  multiplication  is  an  abbrevated  form  of  addition,  and  division, 
of  subtraction,  it  follows  that  every  change  made  upon  the  value  of  a 
number,  must  increase  or  diminish  it.     Hence,  strictly  speaking, 
there  are  but  two  fundamental  operations,  viz. :  aggregation  and 
diminution,  or  increase  and  decrease. 

3.  The  following  problems,  though  subordinate,  are  so  closely  con- 
nected with  the  preceding,  that  a  passing  notice  of  them  may  not  be 
improper,  in  this  connection. 

91.  To  find  the  greater  of  two  numbers,  the  less  and  their 
difference  being  given. 

1.  A  planter  raised  two  successive  crops  of  cotton,  the 
smaller  of  which  amounted  to  4168  bales,  and  the  dif- 
ference between  them  was   1123  bales:   what  was   the 
greater  crop  ? 

ANALYSIS. — If  the  difference  between  two  numbers  be  added  to 
the  less,  it  is  obvious  the  sum  must  be  equal  to  the  greater.  There- 
fore, 4168  bales  +  1123  bales  or  5291  bales  must  be  the  greater  crop. 
Hence,  the 

RULE. — To  the  less  add  the  difference,  and  the  sum  will 
be  the  greater.  (Art.  39.) 

2.  One  of  the  two  candidates  at  a  certain  election,  re- 
ceived 746  votes,  and  was  defeated  by  a  majority  of  411: 
how  many  votes  did  the  successful  candidate  receive  ? 

90.  What  do  these  constitute?  Note.  Why  so  called?  What  is  given  and 
•what  required  in  addition  ?  In  subtraction  ?  In  multiplication  ?  In  division  ? 
Where  begin  the  operation  in  addition,  subtraction,  and  multiplication  ?  Where 
In  division  ?  What  is  the  difference  between  addition  and  subtraction  ?  Between 
addition  and  multiplication?  Between  subtraction  and  division?  Between 
multiplication  and  division  ? 


74  PROBLEMS    AND    FOBMULAS. 

92.  To  find   the  less  of  two   numbers,  the  greater   and 
their  difference  being  given. 

3.  The  greater  of  two  cargoes  of  flour  is  5267  barrels, 
and  their  difference  is  1348  barrels:  how  many  barrels  does 
the  smaller  contain  ? 

ANALYSIS. — The  difference  added  to  the  less  number  equals  the 
greater  ;  therefore,  the  greater  diminished  by  the  difference,  must  be 
equal  to  the  less  ;  and  5267  barrels  minus  1348  barrels  leaves  3919 
barrels,  the  smaller  cargo.  Hence,  the 

RULE. — From  the  greater  subtract  the  difference,  and  the 
result  will  be  the  less.  (Art.  38.) 

4.  At  a  certain  election  one  of  the  two  candidates  re- 
ceived 1366  votes,  and  was  elected  by  a  majority  of  219 
yotes :  how  many  votes  did  the  other  candidate  receive  ? 

93.  The   Product   and   one    Factor   being  given,  to   find  the 

other  Factor. 

5.  A  drover  being  asked  how  many  animals  he  had, 
replied  that  he  had  67  oxen,  and  if  his  oxen  were  multi- 
plied by  his  number  of  sheep,  the  product  would  be 
37520:  how  many  sheep  had  he? 

ANALYSIS. — Since  37520  is  &  product,  and  67  one  of  its  factors,  the 
other  factor  must  be  as  many  as  there  are  673  in  37520 ;  and 
37520-7-67=560.  (Art.  93.)  Therefore,  he  had  560  sheep.  Hence,  the 

EULE. — Divide  the  product  by  the  given  factor,  and  the 
quotient  will  be  the  factor  required.  (Art.  62.) 

6.  The  length  of  a  certain  park  is  320  rods,  and  the 
product  of  its  length  and  breadth  is  5 1 200  rods :  what  is 
its  breadth  ? 

94.  The   Product    of  three    or    more    Factors    and    all   the 

Factors  but  one  being  given,  to  find  that  Factor. 

7.  The  product  of  the  length,  breadth,  and  height  of 
a  certain  mound  is  62730  feet;  its  length  is  45  feet,  and 
its  breadth  41  feet:  what  is  its  height? 

91.  How  find  the  greater  of  two  numbers,  the  less  and  difference  being  given  ? 
93.  How  find  the  less,  the  greater  and  difference  being  given  ? 


PROBLEMS    AND    FORMULAS.  75 

ANALYSIS. — The  contents  of  solid  bodies  are  found  by  multiplying 
their  length,  breadth,  and  thickness  together.  Now  as  the  length  is 
45  feet,  and  the  breadth  41  feet,  the  product  of  which  is  1845,  the 
height  must  be  62730  feet  divided  by  1845,  or  34  feet.  Hence,  the 

RULE. — Divide  the  given  product  by  the  product  of 
the  given  factors,  and  the  quotient  will  be  the  factor  re- 
quired. (Art.  93.) 

8.  The  continued  product  of  the  distances  wK  ;h  4  men 
traveled  is  1944630  miles;    one  traveled  45,  .nother  41, 
and  another  34  miles:  how  far  did  the  fourth  travel? 

95.  To  find  the  Dividend,  the   Divisor  and   Quotient  being 

given. 

9.  If  the  quotient  is  7071,  and  the  divisor  556,  what  is 
the  dividend  ? 

ANALYSIS. — Since  the  quotient  shows  how  many  times  the  cficisor 
is  contained  in  the  dividend,  it  follows,  that  the  product  of  the 
divisor  and  quotient  must  be  equal  to  the  dividend.  Now  7071  x  556 
=3931476  the  dividend.  Hence,  the 

RULE. — Multiply  the  divisor  by  the  quotient,  and  the 
result  will  be  the  dividend.  (Art.  74.) 

10.  "What  number  of  dollars  divided  among  135  persons 
will  give  them  $168  apiece? 

96.  To  find   the   Divisor,   the    Dividend   and   Quotient 

being  given. 

11.  What  must  8640  be  divided  by  that  the  quotient 
may  be  144? 

ANALYSIS. — Since  the  quotient  shows  how  many  times  the  divisor 
is  contained  in  the  dividend,  it  follows  that  if  the  dividend  is  divided 
by  the  quotient  the  result  must  be  the  divisor,  and  8640-*- 144=60. 
Therefore,  the  divisor  is  60.  Hence,  the 

RULE. — Divide  the  dividend  by  tb^  quotient,  and  the 
result  will  be  the  divisor.  (Arts.  62,  93.) 

93.  The  product  and  one  fiactor  being  given,  how  find  the  other  factor  T 


76  PROBLEMS    AND     FOBMULAS. 

12.  If  the  dividend  is  7620,  and  the  quotient  127,  what 
must  be  the  divisor  ? 

97.  The  Sum  and  Difference  of  two  numbers  being  given, 
to  find  the  Numbers. 

13.  The  sum  of  two  numbers  is  65,  and  their  difference 
15  :  what  are  the  numbers  ? 

ANALYSIS. — The  sum  65  is  equal  to  the  greater  number  increased 
by  the  less;  and  the  greater  diminished  by  the  difference  15,  is 
equal  to  the  less  (Art.  38).  Hence,  if  15  is  taken  from  65,  the  re- 
mainder 50,  must  be  twice  the  less.  But  50-5-2=25  the  less  num- 
ber ;  and  25  + 15  =40  the  greater  number. 

Proof. — 40  +  25=65  the  given  sum.    Hence,  the 

KULE. — From  the  sum  take  the  difference,  and  half  the 
remainder  will  be  the  less  number.' 

To  the  less  add  the  difference,  and  the  result  will  be  the 
greater.  (Art.  39.) 

14.  A  merchant  made  $5368  in  two  years,  and  the  dif- 
ference in  his  annual  gain  was  6976 :  what  was  his  profit 
each  year  ? 

15.  The  whole  number  of  votes  cast  for  the  two  candi- 
dates at  a  certain  election  was  5564,  and  the  successful 
candidate  was  elected  by  a  majority  708 :  how  many  votes 
did  each  receive  ? 

16.  A  lady  paid  $250  for  her  watch  and  chain;   the 
former  being  valued  $42  higher  than  the  latter :  what  was 
the  price  of  each  ? 

17.  Two  pupils  A  and  B, solved  75  examples;  B  solving 
15  less  than  A:  how  many  did  each  solve? 

1 3.  A  and  B  found  a  pocket-book,  and  returning  it  to 
the  owner,  received  a  reward  of  $500,  of  which  A  took 
$38  more  than  B :  what  was  the  share  of  each  ? 

94.  Wlien  the  product  of  three  or  more  factors,  and  all  but  one  are  given,  bow 
find  that  one  ?  95.  How  find  the  dividend,  the  divisor  and  quotient  being  given? 
96.  How  find  the  divisor,  the  dividend  and  quotient  being  given  ? 


ANALYSIS. 

98.  Analysis  primarily  denotes  the  separation  of  an 
object  into  its  elements. 

99.  Analysis,  in  Arithmetic,  is  fas  process  of  tracing 
the  relation  of  the  conditions  of  a  problem  to  each  other, 
and  thence  deducing  the  successive  steps  necessary  for  its 
solution. 

NOTES. — i.  The  application  of  Analysis  to  arithmetic  is  of  recent 
date,  and  to  this  source  the  late  improvements  in  the  mode  of 
teaching  the  subject  are  chiefly  due.  Previous  to  this,  Arithmetic 
to  most  pupils,  was  a  hidden  mystery,  regarded  as  beyond  the  reach 
of  all  but  the  favored  few. 

2.  The  pupil  has  already  learned  to  analyze  particular  examples, 
and  from  them  to  deduce  specific  rules  by  which  similar  examples 
may  be  solved  ;  but  Arithmetical  Analysis  has  a  much  wider  range. 
It  is  applied  with  advantage  to  those  classes  of  examples  commonly 
placed  under  the  heads  of  Barter,  Percentage,  Profit  and  Loss, 
Simple  and  Compound  Proportion,  Partnership,  etc.  In  a  word,  it 
is  the  grand  Common-Sense  Rule  by  which  business  men  perform 
the  great  majority  of  commercial  calculations. 

100.  No  specific  directions  can  be  prescribed  for  analyti- 
cal solutions.    The  following  suggestions  may,  however 
be  serviceable  to  beginners : 

ist.  In  general,  we  reason  from  the  given  value  of  one, 
to  the  value  of  two  or  more  of  the  same  kind.  Or, 

2d.  From  the  given  value  of  two  or  more,  to  that  of  one. 

In  the  first  instance  we  reason  from  a  part  to  the 
wJiole  ;  in  the  second,  from  the  whole  to  a  part. 

3d.  Sometimes  the  result  of  certain  combinations  ia 
given,  to  find  the  original  number  or  base. 

98.  What  is  the  primary  meaning  of  analysis  ?  99.  What  is  arithmetical 
analysis?  Note.  What  ia  said  as  to  its  utility  in  arithmetical  and  business 
calculations  ?  100.  Can  specific  rules  be  given  for  analytical  solutions  ?  What 
general  directions  can  you  mention  ? 


78  ANALYSIS. 

In  such  cases,  it  is  generally  best  to  begin  with  the 
result,  and  reverse  each  operation  in  succession,  till  the 
original  number  is  reached.  That  is,  to  reason  from  the 
result  to  its  origin,  or  from  effect  to  cause. 

1.  A  farmer  bought  150  sheep  at  $2  a  head,  and  paid 
for  them  in  cows  at  $20  a  head :  how  many  cows  did  it 
take  to  pay  for  the  sheep  ? 

2.  How  much  tea,  at  85  cents  a  pound,  must  be  given 
for  425  pounds  of  rice,  at  10  cents  a  pound  ? 

3.  How  much  sugar,  at  12  cents  a  pound,  must  be  given 
for  288  pounds  of  raisins,  at  18  cents  a  pound? 

4.  How  much  corn,  at  80  cents  a  bushel,  must  be  given 
for  1 60  pounds  of  tobacco,  at  30  cents  a  pound? 

5.  How  much  butter,  at  40  cents  a  pound,  must  be 
given  for  62  yards  of  calico,  at  20  cents  a  yard  ? 

6.  Bought  189  yards  of  linen,  at  84  cents  a  yard,  and 
paid  for  it  in  oats,  at  42  cents  a  bushel :  how  many  bushels 
did  it  take  ? 

7.  Paid  1 8  barrels  of  flour  for  30  yards  of  cloth,  worth 
$6  a  yard :  what  was  the  flour  a  barrel  ? 

8.  A  farmer  gave  15  loads  of  hay  for  45  tons  of  coal, 
worth  $6  a  ton :  what  did  he  receive  a  load  for  his  hay  ? 

9.  A  sold  B  35  hundred  pounds  of  hops,  at  $27  a  hun- 
dred, and  took  45  sacks  of  coffee,  at  $14  a  sack,  and  the 
balance  in  money :  how  much  money  did  he  receive  ? 

10.  James  bought  96  apples,  at  the  rate  of  4  for  3  cents, 
and  exchanged   them  for  pears  at  4  cents  apiece:   how 
many  pears  did  he  receive  ? 

11.  A  farmer  being  asked  how  many  acres  he  had, 
replied,  if  you  subtract  20  from  the  number,  divide  the 
remainder  by  8,  add  15  to  the  quotient,  and  multiply  by  5, 
the  product  will  be  125  :  how  many  had  he  ? 

ANALYSIS. — Taking  125  as  the  base,  and  reversing  the  several 
operations,  beginning  with  the  last,  we  have  125  -f-  5  =  25,  the 
number  before  the  multiplication.  Again,  subtracting  15  from  25., 
we  have  25  —  15  =  10,  the  number  before  the  addition. 


ANALYSIS.  79 

Next,  multiplying  10  by  8,  we  have  10  x  8=80,  the  number  before 
the  division.  Finally,  adding  20  to  80,  we  have  80  +  20=100,  the 
number  required. 

12.  What  number  is  that,  to  which,  if  25  be  added,  and 
the  sum  multiplied  by  9,  the  product  will  be  504  ? 

13.  What  number,  if  diminished  by  40,  and  the  remain- 
der divided  by  8,  the  quotient  will  be  58  ? 

14.  A  man  being  asked  how  many  children  he  had, 
answered,  if  you  multiply  the  number  by  1 1,  add  23  to 
the  product,  and  divide  the  sum  by  9,  the  quotient  will 
be  16:  how  many  children  had  he? 

15.  The  greater  of  two  numbers  is  3  times  the  less,  and 
the  sum  of  the  numbers  is  36 :  what  are  the  numbers  ? 

ANALYSIS. — The  smaller  number  is  i  part,  and  the  larger  3  parts ; 
hence,  the  sum  of  the  two  is  4  parts,  which  by  the  conditions  is  36. 
Now,  if  4  parts  of  a  number  are  36,  i  part  is  equal  to  as  many  units 
as  there  are  43  in  36,  or  9.  Therefore  9  is  the  smaller  number,  and 
3  times  9  or  27,  the  greater. 

16.  The  sum  of  two  numbers  is  72,  and  the  greater  is 
5  times  the  less*  what  are  the  numbers? 

17.  Divide  472  into  three  such  parts,  that  the  second 
shall  be  twice  the  first,  and  the  third  3  times  the  second 
plus  13. 

ANALYSIS. — Calling  the  first  i  part,  the  second  will  be  2  parts, 
and  the  third  6  parts  plus  13 ;  hence  the  sum  of  the  three,  in  the 
terms  of  the  first,  is  9  parts  plus  13,  which  by  the  conditions  is  472. 
Taking  13  from  472  leaves  459,  and  we  have  9  parts  equal  to  459. 
Now  if  9  parts  equal  459,  I  part  is  equal  to  as  many  units  as  9  is 
contained  times  in  459,  or  51.  Therefore  the  first  is  51,  the  second 
2  times  51  or  102,  the  third  3  times  102  plus  13  or  319. 

1 8.  A  and  B  counting  their  money,  found  that  both 
had  $473,  and  that  A  bad  3  times  as  much  as  B  plus  $25  : 
how  much  had  each  ? 

19.  The   sum  of  two  numbers  is  243,  the  second  is 
three  times  the  first  minus  25  :  what  are  the  numbers? 

20.  What  number  is  that  to  which  if  315  be  added,  the 
sum  will  be  250  less  than  2683  ? 

(For  further  applications  of  Analysis,  see  subsequent  pages.) 


CLASSIFICATION 

AND   PROPERTIES  OF  NUMBERS. 

101.  Numbers  are  divided  into  abstract  and  concrete, 
simple  and  compound,  prime  and  composite,  odd  and  even, 
integral,  fractional,  and  mixed,  known  and  unknown, 
similar  and  dissimilar,  commensurable  and  incommen- 
surable, rational  and  irrational  or  surds. 

DEP. — i.  Abstract  Numbers  are  those  which  are  not  applied  to 
things ;  as,  one,  two,  three. 

2.  Concrete  Numbers  are  those  which  are  applied  to  things ;  as, 
two  caps,  three  pencils,  six  yards. 

3.  Simple  Numbers  are  those  which  contain  only  one  denomination, 
and  may  be  either  abstract  or  concrete;  as,  13,  n  pounds. 

4.  A  Compound  Number  is  one  containing  two  or  more  denomina- 
tions, which  have  the  same  base  or  nature ;  as,  3  shillings  and  6 
pence  ;  4  yards  2  feet  and  6  inches. 

5.  A  Prime  Number  is  one  which  cannot  be  produced  by  multi* 
plication  of  any  two  or  more  numbers,  except  a  unit  and  itself. 

All  prime  numbers  except  2  and  5,  end  in  i,  3,  7,  or  9. 

6.  A  Composite  Number  is  the  product  of  two  or  more  factors,  each 
of  which  is  greater  than  I ;  as,  15  (5  x  3),  24  (2  x  3  x  4),  etc.    (Art.  55.) 

A  prime  number  differs  from  a  composite  number  in  two  respects :  First,  In 
Its  origin ;  Second,  In  its  divisibility  ;  the  former  being  divisible  only  by  a  unit 
and  itself;  the  latter  by  each  of  the  factors  which  produce  it. 

7.  Two  numbers  are  prime  to  each  other  or  relatively  prime,  when 
the  only  number  by  which  both  can  be  divided  without  a  remainder, 
is  a  unit  or  i ;  as,  5  and  6. 

8.  An  Even  Number  is  one  which  can  be  divided  by  2  without  a 
remainder ;  as,  4,  6,  10. 

9.  An  Odd  Number  is  one  which  cannot  be  divided  by  2  without  a 
remainder ;  as,  3,  5,  7,  9,  n. 

Name  the  kind  of  each  of  the  following  numbers,  and  why  :  3,  10, 
17,  21,  28,  31,  56,  63,  72,  81,  44,  39,  91,  67,  51,  84,  99,  100. 

10.  An  Integer  is  a  number  which  contains  one  or  more  entire 
units  only;  as,  i.  3,  7,  10,  50,  100. 

101.  How  are  numbers  divided  ?  An  abstract  number  ?  Concrete  ?  Simple  ? 
Compound?  Prime?  Composite?  When  are  two  numbers  prime  to  each  other ! 
Even?  Odd?  An  integer?  A  fraction?  A  mixed  number? 


PROPERTIES    OF    NUMBERS.  81 

n.  A  Fraction  is  one  or  more  of  the  equal  parts  into  which  a  unit 
is  divided  ;  as,  i-half,  2-thirds,  3-fourths,  etc. 

12.  A  Mixed  Number  is  an  integer  and  a  fraction  expressed 
together  ;  as,  5^,  nf,  etc. 

13.  Like  or  similar  numbers  are  those  which  express  units  of  the 
Bame  kind  or  denomination  ;  as,  3  shillings  and  5  shillings,  four  and 
Beven,  etc. 

14.  Unlike  Numbers  are  those  which  express  units  of  different 
kinds  or  denominations ;  as,  2  apples  and  3  oranges. 

15.  Commensurable  Numbers  are  those  which  can  be  divided  by 
the  same  number,  without  a  remainder ;  as,  9  and  12,  each  of  which 
can  be  divided  by  3. 

1 6.  Incommensurable  Numbers  are  those  which  cannot  be  divided 
by  the  same  number  without  a  remainder.    Thus,  3  and  7  are  incom- 
mensurable. 

17.  A  given  number  is  one  whose  value  is  expressed. 

A  number  is  also  said  to  be  given  when  it  can  be  easily  inferred 
from  something  else  which  is  given.  Thus,  if  two  numbers  are 
given,  their  sum  and  difference  are  given. 

18.  An  Unknown  Number  is  one  whose  value  is  not  given. 

102.  A  Factor  of  a  number  is  one  of  the  numbers, 
which  multiplied  together,  produce  that  number. 

103.  An  Exact  Divisor  of  a  number  is  one  which 
will  divide  it  without  a  remainder.    Thus  2  is  an  exact 
divisor  of  6,,  3  of  15,  etc. 

NOTES. — i.  An  exact  divisor  of  a  nutnbor  is  always  &  factor  of  that 
number ;  and,  conversely,  a  factor  of  a  number  is  always  an  exact 
(Ucisor  of  it.  For,  the  dividend  is  the  product  of  the  duisor  and 
quotient,  and  therefore  is  divisible  by  each  of  the  numbers  that  pro- 
duce it.  (Art.  62.) 

2.  The  terms  dicisor  and  factor  aro  here  restricted  to  integral 
numbers. 

104.  A  Pleasure  of  a  number  is  an  exact  divisor  of 
that  number.     It  is  so  called  because  the  comparative 
magnitude  of  the  number  divided,  is  determined  by  tl.'is 
standard. 

Like  numbers?  Unlike?  Commensurable?  Incommensurable?  What  is  a 
given  number?  An  unknown?  A  tactor?  An  exact  divisor ?  A  measure? 


PROPERTIES     OF     NUMBERS. 

105.  An  aliquot  part  of  a  number  is  a  factor  or  an 
exact  divisor  of  that  number. 

NOTE. — The  terms  factors,  measures,  exact  divisors,  and  aliquot 
parts,  are  different  names  of  the  same  thing,  and  are  often  used  na 
synonymous.  Thus,  3  and  5  are  respectively  the  factors,  measures, 
exact  divisors,  and  aliquot  parts  of  15. 

106.  The  reciprocal  of  a  number  is  i  divided  by 
that  number.    Thus,  the  reciprocal  of  4  is  1-7-4,  or  {. 


COMPLEMENT    OF    NUMBERS. 

107.  The  Complement  of  a  Number  is  the  dif- 
ference between  the  number  and  a  unit  of  the  next  higher 
order.    Thus,  the  complement  of  7  is  3;   for  10  —  7=3; 
the  complement  of  85  is  15  ;  for  100  —  85  —  15. 

108.  To  find  the  Complement  of  a  Number. 

Subtract  the  given  number  from  i  ivith  as  many  ciphers 
annexed,  as  there  are  integral  figures  in  the  given  number. 

Or,  begin  at  the  left  hand,  and  subtract  each  figure  of  the 
given  number  from  9,  except  the  last  significant  figure  on 
th-e  right,  which  must  be  taken  from  10. 

NOTE. — The  second  method  is  based  npon  the  principle  that  when 
we  borrow,  the  next  upper  figure  must  be  considered  i  less  than  it  is. 
(Art.  37,  ?z.) 

Find  the  complement  of  the  following  numbers : 


1.  328. 

6.  6072. 

ii.  56239. 

16.  73245. 

2.  567. 

7.  8256. 

12.  64123. 

17.  1234567. 

3.  604. 

8.  9061. 

13.  102345. 

18.  2301206. 

4.  891. 

9.  13926. 

14.  261436. 

19.  3021238. 

5.  4638. 

10.  23184. 

15.  40061. 

20.  7830426. 

toe,.  An  aliquot  part?  Note.  What  is  said  as  to  the  use  of  these  four  term*T 
106.  The  reciprocal  of  a  number?  107.  The  complement?  108.  How  find  the 
complement  of  a  number  T 


^ 

PROPERTIES    OF    LUMBERS.  83 

/—  / 

DIVISIBILITY    OF    NUMBERS. 

109.  One  number  is  said  to  be  divisible  by  another,     I  .... 
when  there  is  no  remainder.    The  division  is  then  com- 
plete. 

When  there  is  a  remainder,  the  division  is  incomplete; 
and  the  dividend  is  said  to  be  indivisible  by  the  divisor. 

NOTES.  —  i.  In  treating  of  the  divisibility  of  numbers,  the  term 
divisor  is  commonly  used  for  exact  divisor. 

2.  Every  integral  number  is  divisible  by  the  unit  i,  and  by  itself. 
It  is  not  customary,  however,  to  consider  the  unit  i,  or  the  number 
itself,  as  a  factor.  (Art.  102.) 

110.  In  determining  the  divisibility  of  numbers  the 
following  properties  or  facts  are  useful: 

PROP.  i.  Any  number  is  divisible  by  2,  which  ends  with  o,  2,  4, 
6,  or  8. 

2.  Any  number  is  divisible  by  3,  if  the  sum  of  its  digits  is  divisible 
by  3.    Thus,  147  the  sum  of  whose  digits  is  1+4  +  7=12,  is  divisible 
by  3- 

3.  Any  number  is  divisible  by  4,  if  its  too  right  hand  figures  are 
divisible  by  4  ;  as,  2256,  15368.  19384. 

4.  Any  number  is  divisible  by  5,  whicli  ends  with  o  or  5  ;  as, 
130,  675. 

5.  Any  even  number  is  divisible  by  6,  which  is  divisible  by  3. 
Taus,  1344-1-3=448;  and  1344-^-6=224. 

6.  Any  number  is  divisible  by  8,  if  its  three  right  Jutnd  figures  are 
divisible  by  8  ;  as,  1840,  1688,  25320. 

7.  Any  number  is  divisible  by  10,  which  ends  with  o  ;  by  too,  if  it 
ends  with  oo  ;  by  1000,  if  it  ends  with  ooo,  etc. 

8.  Any  number  is  divisible  by  12  which  is  divisible  by  3  and  4. 

9.  Any  number  divided  by  9  will  leave  the  same  remainder  as  the 
turn  of  its  digits  divided  by  9.    Hence, 

10.  Anv  number  is  divisible  by  9  if  the  sum  of  its  digits  is  divisible 
by  9.     Thus,  the  sum  of  the  digits  of  54378  is  5  +  4  +  3  +  7  +  8,  or  27. 
Now,  as  27  is  divisible  by  9,  we  may  infer  that  54378  is  divisible 


ic..  WT.en  is  one  number  divisible  by  another?  Wben  indivisible  ?    MO.  Whe« 
Is  a  number  divisible  by  2  »    By  3?    By<?    By  5?    By  6?    By  8?    By  10? 


84  PROPERTIES     OF     NUMBERS. 

n.  As  9  is  a  multiple  of  3,  any  number  divisible  by  g,  is  also 
divisible  by  3. 

12.  Any  number  divided  by  n  will  leave  the  same  remainder  as 
the  sum  of  its  digits  in  the  even  places,  taken  from  the*w?re  of  those 
in  the  odd  places,  counting  from  the  right,  the  latter  being  increased 
or  diminished  by   n  or  a  multiple  of  11.     Thus,  the  sum  of  the 
digits  in  the  even  places  of  314567,  viz.,  6  +  4  +  3,  or  13.  is  equal  to 
the  sum  of  the  digits  in  the  odd  places,  viz.,  7  +  5  +  1,  or  13 ;  there- 
fore, 314567  is  divisible  by  u.     Heu.ce, 

13.  Any  number  is  divisible  by  n  when  the  sum  of  its  digits  in 
the  even  places  is  equal  to  the  sum  of  those  in  the  odd  places,  or 
when  their  difference  is  divisible  by  n.    (For  a  demonstration  of  the 
properties  of  9  and  n,  see  Higher  Arithmetic.) 

14.  If  one  number  is  a  divisor  of  another,  the  former  is  also  a 
divisor  of  any  multiple  of  the  latter.  (Art.  103,  n.)  Thus,  2  is  a  divisor 
of  6  ;  it  is  also  a  divisor  of  the  product  of  3  times  6,  of  5  x  6,  and  of  any 
whole  number  of  times  6. 

15.  If  a  number  is  an  exact  divisor  of  each  of  two  numbers,  it  is 
nlsa  an  exact  divisor  of  their  sum,  their  difference,  and  their  product. 
Thus,  3  is  a  divisor  of  9  and   15  respectively  ;  it  is  also  a  divisor  ot 
9  +  15,  or  24;  of  15—9,  or  6;  of  15  xg,  or  135. 

16.  A  composite  number  is  divisible  by  each  of  its  prime  factors,  by 
the  product  of  any  two  or  more  of  them,  and  by  no  other  number. 
Thus,  the  prime  factors  of  30  are  2x3x5.     Now  30  is  divisible  by 
by  2,  3  and  5,  by  2  x  3,  by  2  x  5,  by  3  x  5,  and  by  2  x  3  x  5,  and  by  no 
other  number.    Hance, 

17.  The  least  divisor  of  a  composite  number,  is  a  prime  number. 

18.  An  odd  number  cannot  be  divided  by  an  even  number  .vithout 
a  remainder. 

19.  If  an  odd  number  divides  an  even  number,  it  will  also  divide 
half  of  it. 

20.  If  an  even  number  is  divisible  by  an  odd  number,  it  will  also 
be  divisible  by  double  that  number. 

If  a  number  is  divided  by  9,  to  what  is  the  remainder  equal  1  When  is  a 
number  divisible  by  9?  If  a  number  is  divided  by  u,  to  what  is  the  remainder 
equal  ?  When  is  a  number  divisible  by  1 1  V  If  a  number  is  a  divisor  of  another, 
•what  is  true  of  it  in  regard  to  any  multiple  of  that  number?  If  a  number  is  a 
divisor  of  each  of  two  numbers,  what  is  true  of  it  in  regard  to  their  sum,  dif- 
ference, and  product?  By  what  is  a  composite  number  divisible?  Note.  What 
is  true  of  the  least  divisor  of  every  composite  number?  Is  an  odd  number 
dirisible  by  an  even  ?  If  an  odd  divides  an  even  number,  what  is  true  of  half 
of  it?  If  an  even  number  is  divisible  by  an  odd,  what  is  true  of  it  in  regard  to 
double  that  number  ? 


FACTORING. 

111.  Factors,  we  have  seen,  are  numbers  which  mul- 
tiplied together,  produce  &  product.     (Art.  42.)     Hence, 

112.  Factoring  a  number  is  finding  two  or  more  fac- 
tors, which  multiplied  together, produce  the  number.  Thus, 
the  factors  of  15  are  3  and  5  ;  for,  3  x  5  =  15.    (Art.  56,  n.) 

NOTE. — Every  number  is  divisible  by  itself  and  by  i ;  hence,  if 
multiplied  by  i,  the  product  will  be  the  number  itself.  (Art.  41.) 

But  it  cannot  properly  be  said  that  i  is  &  factor,  nor  that  a  number 
is  a  factor  of  itself.  If  so,  all  numbers  are  composite.  (Art.  109,  n.) 

113.  To  resolve  a  Composite  Number  into  two  Factors. 

Divide  the  number  by  any  exact  divisor  ;  the  divisor  and 
quotient  will  be  factors.  (Art.  62.) 

NOTE. — Every  composite  number  must  have  two  factors  at  least ; 
some  have  three  or  more ;  and  others  may  be  resolved  into  several 
different  pairs  of  factors.  (Art.  101,  Del'.  6.) 

1.  What  are  the  two  factors  of  35  ?     Of  49  ?     Of  12 1  ? 

2.  Name  two  factors  of  45  ?    Of  56  ?    Of  72  ?    Of  108  ? 

114.  To   find   the  Different  Pairs  cf  factors   of   a 

Composite  Number. 

3.  What  are  the  different  pairs  of  factors  of  36  ? 

ANALYSTS. — Dividing  36  by  2,  we  have  36-5-2=18.  36  —  2-=  18 

Again,  36-1-3=12;  36-4-4=9;  and  36-5-6=6.     That  36  —  3=12 

is,  the  different  pairs  of  factors  of  36  are  2x18;  36 —  4=   o 

jx  12;  4x9;  and  6x6.    Hence,  the  36  —  6—   5 

EULE. — Divide  the  given  number  continually  by  each  of 
its  exact  divisors,  beginning  ivith  the  least,  until  the  quo* 
tient  obtained  is  less  than  the  divisor  employed. 

The  divisors  and  corresponding  quotients  will  be  th& 
different  pairs  of  factors  required. 

m.  What  arc  factors?  Is  i  a  factor?  Is  a  number  a  factor  of  itself  ?  Why 
not  ?  What  is  meant  by  Factoring?  Note.  How  resolve  a  number  into  two  fkc. 
tore  ?  1 13.  How  find  the  different  pairs  of  factors  T 


86  FACTORING. 

NOTE. — The  division  is  stopped  as  soon  as  the  quotient  is  lest  than 
the  divisor ;  for,  the  subsequent  factors  will  be  similar  to  those 
already  found. 

Find  the  different  pairs  of  factors  of  the  following 
numbers : 

4-  20,  8.  38,  12.  96,  16.  256, 

5-  27>  9-  45>  13-  1 10,  17.  475, 
6.  30,  10.  56,  14.  144,  18.  600, 
7-  32,  "•  75,  !5-  225,  19.  1240. 


PRIME    FACTORS. 

115.  The  Prime  Factors  of  a  number  are  the 
Prime  Numbers  which,  multiplied  together,  produce  the 
number. 

116.  Every  Composite  Number  can  be  resolved 
into  prime  factors.    For,  the  factors  of  a  composite  num- 
ber are  divisors  of  it,  and  these  divisors  are  either  prime 
or  composite.    Ifjyrime,  they  accord  with  the  proposition. 
If  composite,  they  can  be  resolved  into  other  factors,  and 

on,  until  all  the  factors  are  prime.     (Art.  1 14.) 

117.  To  find  the  Prime  factors  of  a  Composite  Number. 
Ex.  i.  What  are  the  prime  factors  of  210  ? 

ANALYSIS.  —  Dividing    210  OPERATION. 

by  any   prime   number  that        ist  divisor,  2   |    2 1 o,  given. 


will  exactly  divide  it,  aa  2,  2d  "  3 

we  resolve  it  into  the  factors  3d  "  5 

2  and  105.     Again,  dividing  4th  "  7 
the  ist  quotient  105  by  any 


105,  ist  quot. 

35,  2d     " 

7,3d     « 

i,  4th    « 


prime  number,  as   3,  we  re-    Hence,  210  =  2x3x5x7. 
solve  it  into  the  factors  3  and 

35.  In  like  manner,  dividing  the  2d  and  3d  quotients  by  the  prime 
numbers  5  and  7,  the  4th  quotient  is  i.  The  divisors  2,  3,  5  and  7 
are  the  prime  factors  required.  For,  each  of  these  divisors  ia  a 
prime  number,  and  the  division  is  continued  until  the  quotient  is  a 

115.  What  are  prime  factors?     116.  Wliat  is  said  of  composite  numbers? 
How  find  the  prime  factors  of  a  number? 


FACTORING.  87 

unit ;  therefore,  the  several  divisors  must  be  the  prime  factor* 
required.     Hence,  the 

RULE. — Divide  the  given  number  by  any  prime  number 
that  will  divide  it  without  a  remainder.  Again,  divide 
this  quotient  by  a  prime  number,  and  so  on,  until  the  quo- 
tient obtained  is  i.  The  several  divisors  are  the  prime 
factors  required. 

NOTE. — As  the  least  divisor  of  every  number  is  prime,  beginners 
may  be  less  liable  to  mistakes  by  taking  for  the  divisor  the  smallest 
number  that  will  divide  the  several  dividends  without  a  remainder 

Find  the  prime  factors  of  the  following  numbers : 


2. 

72. 

7- 

184. 

12. 

IOOO. 

17- 

2348. 

3- 

96. 

8. 

215- 

13- 

1208. 

1  8. 

10376. 

4- 

121. 

9- 

320. 

14. 

1560. 

19. 

25600. 

5- 

I32. 

10. 

468. 

15- 

1776. 

20. 

64384. 

6. 

144. 

ii. 

576. 

1  6. 

1868. 

21. 

98816. 

118.  To  find  the  Prime  Factors  common  to  two  OP  more 
Numbers. 

22.  What  are  the  prime  factors  common  to  42,  168,  and 
210  ? 

ANALYSIS. — Dividing  the  given  numbers  by  the 

prime  factor  2,  the  quotients  are  21,  84,  and  105.  OPERATION. 

Again,  dividing  these  quotients  by  the  prime  2)42,  168,  210 

factor  3,  the  quotients  are  7,  28,  and  35.     Finally,  ~\^ 84    IO5 

dividing  by  7,  the  quotients  are  i,  4,  and  5,  no  two  r-- 

of  which  can  be  divided  by  any  number  greater  •>    7>     2">     35 

than  i.     Therefore,  the  divisors  2,  3,  and  7  aro  ij       4>       5 
the  common  prime  factors  required.     Hence,  the 

RULE. — I.  Write  the  given  numbers  in  a  horizontal  line, 
and  divide  them  by  any  prime  number  which  will  divide 
each  of  them  without  a  remainder. 

II.  Divide  the  quotients  thence  arising  in  the  same 
manner,  as  long  as  the  quotients  have  a  common  factor ; 

the  divisors  ii'ill  be  the  prime  factors  required. 

~* 

118.  How  find  the  prime  factors  common  to  two  or  more  numbers? 


88  CANCELLATION. 

Find    the    prime   factors    common   to    the 
numbers : 

23.  12,  1 8,  30.  29.  75,  90,  135,  150. 

24.  48,  66,  72.  30.  132,  144,  196,  240. 

25.  64,  108,  132.  31.  168,  256,  320,  500. 

26.  71,  113,  149.  32.  200,  325,  540,  625. 

27.  48,  72,  88,  120.  33.  316,  396,  484,  936. 

28.  60,  84,  108,  144.  34.  462,  786,  924,  858,  972. 

119.  To  find  the  Prime  Numbers  as  far  as  250. 

I.   Write  the  odd  numbers  in  a  series,  including  2. 
II.  After  3,  erase  all  that  are  divisible  by  3  ;  after  5,  erase 
all  that  are  divisible  by  5  ;  after  7,  all  that  are  divisible  by  7  ; 
after  1 1  and  1 3,  all  that  are  divisible  by  them :  those  left 
are  primes. 

35.  Find  what  numbers  less  than  100  are  prime. 

36.  Find  what  numbers  less  than  200  are  prime. 


CANCELLATION. 

120.  Cancellation  is  the  method  of  abbreviating  opera- 
tions by  rejecting  equal  factors  from  the  divisor  and  dividend. 

The  Sign  of  Cancellation  is  an  oblique  mark 
drawn  across  the  face  of  a  figure ;  as,  S,  5, 1,  etc. 
NOTE. — Tlie  term  cancellation  is  from  the  Latin  cancello,  to  erase. 

121.  Cancelling  a  Factor  of  a  number  divides 
the  number  by  that  factor.     For,  multiplying  and  divid- 
ing a  number  by  the  same  factor  does  not  alter  its  value. 
(Art.  86,  a.}   Thus,  let  7  x  5  be  a  dividend.  Now  cancelling 
the  7,  divides  the  dividend  by  7,  and  leaves  5  for  the 
quotient ;  cancelling  the  5,  divides  it  by  5,  and  leaves  7 
for  the  quotient. 

120.  What  is  cancellation  T     121.  Effect  of  cancelling  a  factor?    WhyT 


CANCELLATION".  89 

122.  To  divide  one  Composite  Number  by  another. 

1.  What  is  the  quotient  of  72  divided  by  18  ? 

ANALYSIS.  —  The  dividend  72=9  x  4  x  2,  and  OPERATION. 

the  divisor  18=9x2.     Cancelling  the  com-  72      SI  X  4  X  2 

mon   factors  9  and    2,  we  have   4  for  the  Y8~    Six  2    =^ 
quotient. 

2.  Divide  the  product  of  45  x  17,  by  the  product  of  9  x  5, 

ANALYSIS.  —  In  this  example  the  product  of  the  two  9  45 

factors  composing  the  divisor,  viz.  :   9x5,   equals  one 


' 


factor  of  the  dividend.    We  therefore  cancel  these  equals    ~\ 
at  once,  and  the  factor  17  is  the  quotient. 

Remark.  —  In  this  operation  we  place  the  numbers  composing  the  divisor  on 
the  left  of  a  perpendicular  line,  and  those  composing  the  dividend  on  the  right. 
The  result  is  the  same  as  if  the  divisor  were  placed  under  the  dividend  as  before. 
Each  method  has  its  advantages,  and  may  be  used  at  pleasure.  Hence,  the 

RULE.  —  Cancel  all  the  factors  common  to  the  divisor  and 
dividend,  and  divide  the  product  of  those  remaining  in  the 
dividend  by  the  product  of  thosz  remaining  in  the  divisor. 
(Art  86.) 

NOTES.  —  i.  When  either  the  divisor,  dividend,  or  both,  consist  of 
two  or  more  factors  connected  by  the  sign  x  ,  they  may  be  regarded 
as  composite  numbers,  and  the  common  factors  be  cancelled  before 
the  multiplication  is  performed. 

2.  This  rule  is  founded  upon  the  principle  that,  if  the  divisor  and 
dividend  are  both  divided  by  the  same  number,  the  quotient  is  not 
altered.     (Art.  86.)    Its  utility  is  most  apparent  in  those  operations 
which  involve  both  multiplication  and  division;  as  Fractions,  Pro- 
portion, etc. 

3.  When  the  factor  or  factors  cancelled  equal  the  number  itself, 
the  unit   i,  is  always  left  ;  for,  dividing  a  number  by  itself,  the 
quotient  is  i.     When  the   i   stands  in  the  dividend,  it   must  be 
retained.    But  when  in  the  divisor,  it  is  disregarded.    (Art.  81.) 

3.  Divide  the  product  5  x  6  x  8  by  30  x  48. 

SOLUTION.  —  Cancelling  the  common  5x6x8  5  x  6  x  "8.  i  i 
factors,  we  have  I  in  the  dividend  ~  Q  =  ,»«  jTiT?  6  =  6 

and  6  in  the  divisor.    (Art.  112,  n.) 

.13.  Rule  for  cancelling  in  division  ?    Note.  On  what  is  the  rule  founded  J 


90  CANCELLATION. 

4.  Divide  24  x  6  by  12  x  7.         7.  Divide  18  x  15  by  5  x  6. 

5.  Divide  42  x  8  by  21  x  2.         8.  Divide  56  x  16  by  8  x  4. 

6.  18  X3  X4by  12  x6.  9.  28  x6  x  12  by  14  x  8. 

10.  Divide  21  x  5  x  6  by  7  x  3  x  2. 

1 1.  Divide  32  x  7  x  9  by  8  x  5  x  4. 

12.  Divide  27  x  35  x  i4by  9  x  7  x  21. 

13.  Divide  33  x  42  x  25  by  7  x  5  x  1 1. 

14.  Divide  36  x  48  x  56  x  5  by  96  x  8  x  4  x  2. 

15.  Divide  63  x  24  x  33  x  2  by  9  x  12  x  n. 

16.  Divide  175  x  28  x  72  by  25  x  14  x  12. 

17.  Divide  220  x  60  x  48x69  by  13  x  nox  12  x8. 

18.  Divide  350  x  63  x  144  by  50  x  72  x  24. 

19.  Divide  500  x  128  x  42  x  108  by  256  x  250  x  12. 

20.  How  many  barrels  of  flour,  at  $12  a  barrel,  must  be 
given  for  40  tons  of  coal,  at  $9  per  ton  ? 

21.  How  many  tubs  of  butter  of  56  pounds  each,  at 
28  cents  a  pound,  must  be«given  for  8  pieces  of  shirting, 
containing  45  yards  each,  at  26  cents  a  yard  ? 

22.  How  many  bags  of  coffee  42  pounds  each,  worth 
4  shillings  a  pound,  must  be  given  for  18  chests  of  tea, 
each  containing  72  pounds,  at  6  shillings  a  pound? 

23.  Bought  24  barrels  of  sugar,  each  containing  168 
pounds,  at  20  cents  a  pound,  and  paid  for  it  in  cheeses, 
each  weighing  28  pounds,  worth  16  cents  a  pound:  how 
many  cheeses  did  it  take  ? 

24.  A  grocer  sold  10  hogheads  of  molasses,  each  con- 
taining 63  gallons,  at  7  shillings  per  gallon,  and  took  his 
pay  in  wheat,  worth  14  shillings  a  bushel:  how  much 
wheat  did  he  receive  ? 

25.  A  young  lady  being  asked  her  age,  replied:  If  you 
divide  the  product  of  64  into  14,  by  the  product  of  8  into 
4,  you  will  have  my  age :  what  was  her  age  ? 

26.  A  has  60  acres,  worth  $15  an  acre,  and  B  100  acres, 
worth  $75  an  acre:   B's  property  is  how  many  times  the 
value  of  A's  ? 


COMMON   DIVISORS. 

123.  A  Common  Divisor  is  any  number  that  will 
divide  two  or  more  numbers  without  a  remainder. 

124.  To  find  a  Common  Divisor  of  two  or  more  Numbers. 

1.  Required  a  common  divisor  of  10,  12,  and  14. 

AHALYSIS. — By  inspection,  we  perceive  that  10=  10  =  2x5 

2x5;  12=2x6,  and  14=2x7.     The  factor  2,  is  com-  12  =  2x6 

mon  to  each  of  the  given  numbers,  and  is  therefore  14  =  2  X  7 
a  common  divisor  of  them.  (Art.  123.)  Hence,  the  Ans.  2. 

RULE. — Resolve  each  of  the  given  numbers  into  two 
factors,  one  of  which  is  common  to  them  all. 

Find  common  divisors  of  the  following  numbers : 

2.  8,  1 6,  20,  and  24.  Ans.  2  and  4. 

3.  12,  15,  1 8,  and  30.  6.  21,  28,  35,  49,  63. 

4.  36,  48,  96,  and  108.  7.  20,  30,  70,  and  100. 

5.  42,  54,  66,  and  132.  8.  60,  75,  120,  and  240. 
9.  16,  24,  40,  64,  116,  120,  144,  168,  264,  1728. 

GREATEST    COMMON    DIVISOR. 

125.  The  Greatest  Common  Divisor  of  two  or 

more  numbers,  is  the  greatest  number  that  will  divide 
each  of  them  without  a  remainder.  Thus,  6  is  the  greatest 
common  divisor  of  18,  24,  and  30. 

NOTEa — I.  A  common  divisor  of  two  or  more  numbers  is  the  same 
as  a  common  factor  of  those  numbers ;  and  the  greatest  common 
divisor  of  them  is  their  greatest  common  factor.  Hence, 

2.  If  any  two  numbers  have  not  a  common  factor,  they  cannot 
have  a  common  divisor  greater  than  i.    (Art.  112,  n.) 

3.  A  common  divisor  is  often  called  a  common  measure,  and  the 
greatest  common  divisor,  the  greatest  common  measure. 

123.  A  common  divisor?  125.  Greatest  common  divisor?  Note.  A  common 
divisor  is  the  same  as  what  ?  Often  called  what  ? 


COMMON     DIVISOBS. 


FIRST    METHOD. 

126.  To  find  the  Greatest  Common  Divisor  by  continued 
Divisions. 

i.  What  is  the  greatest  common  divisor  of  28  and  40  ? 

ANALYSIS. — If  we  divide  the  greater  number 
by  the  less,  the  quotient  is  i,  and  12  remainder.          OPERATION. 
Next,  dividing  the  first  divisor  28,  by  the  first      28)40(1 
remainder  12,  the  quotient  is  2,  and  4  remainder.  28 

Again,  dividing  the  second  divisor  by  the  second  ~\    n/ 

remainder  4,  the  quotient  is  3,  and  o  rem.     The 
last  divisor  4,  is  the  greatest  common  divisor. 

Demonstration. — We  wish  to  prove  two  points :  4) 1 2  (3 

ist,  That  4  is  a  common  divisor  of  the  given  num-  -1 2 

bers.    2d,  That  it  is  their  greatest  common  divisor. 

First.  We  are  to  prove  that  4  is  a  common  divisor  of  28  and  40. 
By  the  last  division,  4  is  contained  in  12,  3  times.  Now,  as  4  is  a 
divisor  of  12  ;  it  is  also  a  divisor  of  the  product  of  12  into  2,  or  24. 
(Art.  no,  Prop.  14.)  Next,  since  4  is  a  divisor  of  itself  and  24,  it 
must  be  a  divisor  of  the  sum  of  4  +  24,  or  28,  which  is  the  smaller 
number.  (Prop.  15.)  For  the  same  reason,  since  4  is  a  divisor  of  12 
and  28,  it  must  also  be  a  divisor  of  the  sum  of  12  +  28,  or  40,  which 
is  the  larger  number.  Hence,  4  is  a  common  divisor  of  28  and  40. 

Second.  We  are  now  to  prove  that  4  is  the  greatest  common  divisor 
of  28  and  40.  If  the  greatest  common  divisor  of  these  numbers  is 
not  4,  it  must  be  either  greater  or  less  thau  4.  But  we  have  shown 
that  4  is  a  common  divisor  of  the  given  numbers;  therefore,  no 
number  less  than  4  can  be  the  greatest  common  divisor  of  them. 
The  assumed  number  must  therefore  be  greater  than  4.  By  supposi- 
tion, this  assumed  number  is  a  divisor  of  28  and  40;  hence,  it  must 
be  a  divisor  of  their  difference  40—28  or  12.  And  as  it  is  a  divisor  of 
12,  it  must  also  divide  the  product  of  12  into  2  or  24.  Again,  since 
the  assumed  number  is  a  divisor  of  28  and  24,  it  must  also  be  a  di- 
visor of  their  difference,  which  is  4 ;  that  is,  a  greater  number  will 
divide  a  less  without  a  remainder,  which  is  impossible.  (Art.  81, 
Prin.  2.)  Therefore,  4  must  be  the  greatest  common  divisor  of  28 
and  40.  Hence,  the 

126.  Explain  Ex.  i.  Prove  that  4  is  the  greatest  common  divisor  of  28  and  40. 
Enle  i  Note.  If  there  are  more  thau  two  numbers,  how  proceed  ?  If  tb-e  Uet 
divisor  is  i,  then  what?  What  is  the  greatest  common  divisor  of  two  or 
prime  numbers,  or  numbers  prime  to  each  other  ? 


COMMON    DIVISOES.  93 

RULE. — Divide  the  greater  number  by  the  less;  then 
divide  the  first  divisor  by  the  first  remainder,  the  second 
divisor  by  the  second  remainder,  and  so  on,  until  nothing 
remains;  the  last  divisor  will  be  the  greatest  common 
divisor. 

NOTES. — i.  If  there  are  more  than  two  numbers,  begin  with  the 
smaller,  and  find  the  greatest  common  divisor  of  two  of  them ;  then 
of  this  divisor  and  a  third  number,  and  so  on,  until  all  the  numbers 
have  been  taken. 

2.  The  greatest  common  divisor  of  two  or  more  prime  numbers, 
or  numbers  prime  to  each  other,  is  i.    (Art.  112,  n.) 

3.  If  the  last  divisor  is  i,  the  numbers  are  prime,  or  pri-me  to  each 
other  ;  therefore  their  greatest  common  divisor  is  i.     Such  numbers 
are  said  to  be  incommensurable.    (Art.  101,  Def.  16.) 

2.  What  is  the  greatest  com,  divisor  of  48,  72,  and  108  ? 

3.  What  is  the  greatest  common  divisor  of  72  and  120? 

SECOND    METHOD. 

127.    To   find    the    greatest   Common  Divisor    by  Prime 
Factors. 

4.  What  is  the  greatest  common  divisor  of  28  and  40  ? 
ANALYSIS. — Resolving  the  given  numbers  into 

OPERATION. 

factors   common  to  both,  we  have  28=2x2x7,        0 

XT          *i.  j  f  ^i.  2o  =  2  X  2  X  7 

and  40=2x2x10.    Now  the  product  of  these 

factors,   viz.,  2  into  2,  gives  4  for  the  greatest      4O  =  2  x  2  x  IO 

common  divisor,  the  same  as  by  the  first  method.     Ans.  2x2  =  4. 

5.  What  is  the  greatest  com.  div.  of  30,  45,  and  105  ? 

ANALYSIS. — Setting  the  numbers  in  a  hori-  OEEKATIOH. 

zontal  line,  we  divide  by  any  prime  number,  as          3)30?  45?  l°5 
3,  that  will  divide  each  without  a  remainder,  e^io    T<      T? 

and  set  the  quotients  under  the  corresponding 
numbers.     Again,  -dividing  each  of  these  quo-  '     *» 

tieuts  by  the  prime  number  5,  the  new  quotients     3  *  5  ~  *5> 
2,  3,  and  7,  are  prime,  and  have   no  common 
factor.     Therefore,  the  product  of  the  common  divisors  3  into  5,  or 
15,  is  the  gxaatest  common  divisor.    (An.  no,  Prop.  16)    Hence,  the 

127.  What  is  the  rule  for  the  second  method?  Note.  What  is  the  object  of 
placing  the  numbers  in  a  horizontal  line  ? 


COMMON    DIVISORS. 

RULE. — I.  Write  the  numbers  in  a  horizontal  line,  and 
divide  by  any  prime  number  that  will  divide  each  without 
a  remainder,  setting  the  quotients  in  a  line  below. 

II.  Divide  these  quotients  as  before,  and  thus  proceed  till 
*io  number  can  be  found  that  will  divide  all  the  quotients 
without  a  remainder.  The  product  of  all  the  divisors  will 
be  the  greatest  common  divisor. 

NOTES. — i.  This  rule  is  the  same  as  resolving  the  given  numbers 
into  prime  factors,  and  multiplying  together  all  that  are  common. 

2.  The  advantage  of  placing  the  numbers  in  a  horizontal  line,  is 
that  the  prime  factors  that  are  common,  may  be  seen  at  a  glance. 

3.  When  required  to  find  the  greatest  Common  divisor  of  three  or 
•more  numbers,  the  second  method  is  generally  more  expeditious,  and 
therefore  preferable. 

4.  When  the  given  numbers  have  only  one  common  factor,  that 
factor  is  their  greatest  common  divisor. 

5.  Two  or  more  numbers  may  have  several  common  divisors ;  but 
they  can  have  only  one  greatest  common  divisor. 

Find  the  greatest  common  divisor  of  the  following: 

6.  63  and  42.  14.  10,  28,  40,  64,  90,  32. 

7.  135  and  105.  15.  12,  36,  60,  108,  132. 

8.  24,  36,  72.  16.  16,  28,  64,  56,  160,  250. 

9.  60,  75,  12.  17.  576  and  960. 

10.  75,  125,  250.  18.   1225  and  592. 

11.  42,  54,  60,  84.  19.  703  and  1369. 

12.  72,  ioo,  168,  136.  20.   1492  and  1866. 

13.  60,  84,  132,  108.  21.  2040  and  4080. 

22.  A  merchant  had  180  yards  of  silk,  and  234  yards 
of  poplin,  which  he  wished  to  cut  into  equal  dress  pat- 
terns, each  containing  the  greatest  possible  number  of 
yards :  how  many  yards  would  each  contain  ? 

23.  Two  lads  had  42  and  63  apples  respectively :  how 
many  can  they  put  in  a  pile,  that  the  piles  shall  be  equal, 
and  each  pile  have  the  greatest  possible  number  ? 

24.  A  man  had  farms  of  56,  72,  and  88  acres  respectively, 
which  he  fenced  into  the  largest  possible  fields  of  the  same 
number  of  acres :  how  many  acres  did  he  put  in  each  ? 


MULTIPLES. 


Multiple  is  a  number  which  can  be  dividec 
by  another  number,  without  a  remainder.  Thus,  12  is  a 
multiple  of  4. 

REMAKK. — The  Term  Multiple  is  also  used  in  the  sense 
of  product :  as  when  it  is  said,  "  if  one  number  is  a  di- 
visor of  another,  the  former  is  also  a  divisor  of  any  mul- 
tiple (product)  of  the  latter."  Thus,  3  is  a  divisor  of  6 ; 
it  is  also  a  divisor  of  7  times  6,  or  42. 

NOTE. — Multiple  is  from  the  Latin  multiplex,  having  many  folds, 
or  taken  many  times  ;  hence,  a  product, 

But  every  product  is  divisible  by  its  factors;  hence,  the  terra  came 
to  denote  a  dividend.  The  former  signification  is  derived  from  the 
formation  of  the  number ;  the  latter,  from  its  divisibility. 

129.  A  Common  Multiple  is  any  number  that 
can  be  divided  by  two  or  more  numbers  without  a  re- 
mainder.   Thus,  1 8  is  a  common  multiple  of  2,  3,  6,  and  9. 

130.  A  common  multiple  of  two  or  more  numbers  may 
be  found  by  multiplying  them  together.    That  is,  the  pro- 
duct of  two  numbers,  or  any  entire  number  of  times  their 
product,  is  a  common  multiple  of  them.     (Art.  128,  n.) 

NOTES. — i.  The  factors  or  divisors  of  a  multiple  are  sometimes 
called  sub-multiples. 

2.  A  number  may  have  an  unlimited  number  of  multiples.  For, 
according  to  the  second  definition,  every  number  is  a  multiple  of 
itself;  and  if  multiplied  by  2,  the  product  will  be  a  second  multiple ; 
if  multiplied  by  3,  the  product  will  be  a  third  multiple ;  and  univer- 
sally, its  product  into  any  ichole  number  will  be  a  multiple  of  that 
number.  (Art.  128.) 

Find  a  common  multiple  of  the  following  numbers: 

1.  2,  3  and    5.          4.     2,    8  and  10.          7.  13,    7  and  22. 

2.  3,  7  and  n.          5.     7,    6  and  17.          8.  19,    2  and  40. 

3.  5,  7  and  13.         6.  n,  21  and  31.          9.   17,  10  and  34. 

128.  Meaning  of  the  term  multiple?  129.  What  is  a  common  multiple? 
130.  How  found?  Not*.  \Vhat  are  sub-multiples ?  How  many  multiples  has  a 
number  T 


96  MULTIPLES. 


LEAST    COMMON    MULTIPLE. 

131.  The  Least  Comm-on  Multiple  of  two  or 

more  numbers,  is  the  least  number  that  can  be  divided 
by  each  of  them  without  a  remainder.  Thus,  15  13  the 
least  common  multiple  of  3  and  5. 

132.  A  Multiple,  according  to  the  first  definition,  is  a 
composite  number.     But  a  composite  number  contains  all 
the  prime  factors  of  each  of  the  numbers  which  produce 
it.    Hence,  we  derive  the  following  Principles  : 


.  i.  That  a  multiple  of  a  number  must  contain  all  the  prime 
factors  of  that  number. 

2.  A  common  multiple  of  two  or  more  numbers  must  contain  all 
the  prime  factors  of  each  of  the  given  numbers. 

3.  The  least  common  multiple  of  two  or  more  numbers  is  the  least 
number  which  contains  all  their  prime  factors,  each  factor  being 
taken  as  many  times  only,  as  it  occurs  in  either  of  the  given  numbers, 
and  no  more. 

133.    To  find  the  Least  Common   IMLultiple  of  two   OP 

more  numbers..  % 

i.  What  is  the  least  common  multiple  of  1  8,  21  and  66  ? 

ist  METHOD.  —  We  write  the  num-  ,6t  OPERATION. 

bera  in  a  horizontal  line,  with  a  comma  2)  18    21    66 

between  them.    Since  2  is  a  prime  fac- 
tor of  one  or  more  of  the  given  num- 

bers, it  must  be  a  factor  of  the  least  3>       7>  * 

common  multiple.  (Art.  132,  Prin.  i.)  2x3x3x7x11  =  13801 
We  therefore  divide  by  it,  setting  the 

quotients  and  undivided  numbers  in  a  line  below.  In  like  manner,  we 
divide  these  quotients  and  undivided  numbers  by  the  prime  number 
3,  and  set  the  results  in  another  line,  as  before.  Now,  as  the  numbers 
in  the  third  line  are  prime,  we  can  carry  the  division  no  further  ; 
for,  they  have  no  common  divisor  greater  than  i.  (Art.  101.) 
Hence,  the  divisors  2  and  3,  with  the  numbers  in  the  last  line, 
3,  7  and  n,  are  all  the  prime  factors  contained  in  the  given  numbers, 
and  each  is  taken  as  many  times  as  it  occurs  in  either  of  them. 

131.  What  !s  the  least  common  multiple  ?    132.  What  principles  are  derived? 


MULTIPLES.  97 

Therefore,  the  continued  product  of  these  factors,  2x3x3x7x11, 
or  1386,  is  the  least  common  multiple  required. 

2d  METHOD. — Resolving  each  num-  ad  OPKRATIOK. 

ber  into  its  prime   factors,  we  have  1 8  =  2  X  3  X    3 

18=2x3x3,  21=3x7,   and  66=2x3  21  =  3x7 

xii.     Now,  as  the  least  common  mul-  66  =2x3x11 

tiple  must  contain  all  the  prime  fac- 

tors  of  the  given  number,  it  must  con-  2X3XIIX7X  3  =  1386 
tain  those  of  66,  viz.,  2x3x11;  we 

therefore  retain  these  factors.  Again,  it  must  contain  the  prime 
factors  of  21,  which  are  3x7,  and  of  18,  which  are  2x3x3,  each 
being  taken  as  many  times  as  it  is  found  in  either  of  the  given 
numbers.  But  2  is  already  retained,  and  3  has  been  taken  once;  we 
therefore  cancel  the  2  and  one  of  the  33,  and  retain  the  other  3  and 
the  7.  The  continued  product  of  these  factors,  viz.,  2x3x11x7x3, 
is  1386,  the  same  as  before.  Hence,  the 

RULE. — I.  Write  the  numbers  in  a  horizontal  line,  and 
divide  by  any  prime  number  that  will  divide  two  or  more 
of  them  without  a  remainder,  placing  the  quotients  and 
numbers  undivided  in  a  line  below. 

II.  Divide  this  line  as  before,  and  thus  proceed  till  no 
two  numbers  are  divisible  by  any  number  greater  than  i. 
The  continued  product  of  the  divisors  and  numbers  in  the 
last  line  will  be  the  anszver. 

Or,  resolve  the  numbers  into  their  prime  factors  ;  mid- 
tiply  these  factors  together,  taking  each  the  greatest  number 
of  times  it  occurs  in  either  of  the  given  numbers.  TJie 
product  will  be  the  answer. 

NOTES. — i.  These  two  methods  are  based  upon  the  same  principle, 
viz. :  that  the  least  common  multiple  of  two  or  more  numbers  is  the 
least  number  which  contains  all  their  prime  factors,  each  factor 
being  taken  as  many  times  only,  as  it  occurs  in  either  of  the  given 
numbers.  (Art.  132,  Prin.  3.) 

2.  The  reason  we  employ  prime  numbers  as  divisors  is  because  the 
given  numbers  are  to  be  resolved  into  prime  factors,  or  factors  prime 

133.  How  find  the  least  common  multiple  ?  Note.  Upon  what  principle  are  these 
two  methods  based  ?  Whr  employ  prime  numbers  for  divisors  in  the  first 
method?  Why  write  the  numbers  in  a  horizontal  line?  What  is  the  second 
method  ?  Advantage  of  it  ?  How  may  the  operation  be  shortened  ?  When  the 
given  numbers  are  prime,  or  prime  to  each  other,  what  U  to  be  done  ? 


98  MULTIPLES. 

to  each  other.  If  the  divisors  were  composite  numbers,  they  would 
be  liable  to  contain  factors  common  to  some  of  the  quotients,  or 
numbers  in  the  last  line ;  and  if  so,  their  continued  product  would 
not  be  the  least  common  multiple. 

3.  The  object  of  placing  the  numbers  in  a  horizontal  line,  is  to 
resolve  all  tho  numbers  into  prime  factors  at  the  same  lime. 

4.  The  chief  advantage  of  the  second  method  lies  in  the  distinct- 
ness with  which  the  prime  factors  are  presented.     The  former  is  the 
more  expeditious  and  less  liable  to  mistakes. 

5.  The  operation  may  often  be  shortened  by  cancelling  any  num- 
ber which  is  a  factor  of  another  number  in  the  same  line.    (Ex.  2.) 
When  the  given  numbers  are  prime,  or  prime  to  each  other,  they 
have  no  common  factors  to  be  rejected ;   consequently,  their  con- 
tinued product  will  be  the  least  common  multiple.     Thus,  the  least 
common  multiple  of  3  and  5  is  15  ;  of  4  and  9  is  36. 

2.  Find  the  least  common  multiple  of  5,  6,  9,  10,  21. 

ANALYSIS. — In  the  first  line  5  2)5,  6,  9,  10,  21 

is  a  factor  of  10,  and  is  therefore  \«  ~    ~~ 

cancelled ;  in  the  second,  3  is  a 
iactor  of  9,  and  is  also  cancelled.  3>     5>     7 

The  product  of  2x3x3x5x  7=630,  the  answer  required. 

Find  the  least  common  multiple  of  the  following : 

3.  8,  12,  16,  24.  10.  36,  48,  72,  96. 

4.  14,  28,  21,  42.  ii.  42,  68,  84,  108. 

5  36,  24,  48,  60.  12.   120,  144,  1 68,  216. 

6.  25,  40,  75,  100.  13.  96,  108,  60,  204. 

7.  1 6,  24,  32,  40.  14.  126,  154,  280,  560. 

6  22,  33,  55,  66.  15.  144,  256,  72,  300. 
9.  30,  40,  60,  80.  16.  250,  500,  1000. 

17.  Investing  the  same  amount  in  each,  what  is  the 
smallest  sum  with  which  I  can  buy  a  whole  number  of 
pears  at  4  cents,  lemons  at  6,  and  oranges  at  10  cents  ? 

1 8.  A  can  hoe  16  rows  of  corn  in  a  day,  B  18,  C  20,  and 
D  24  rows :  what  is  the  smallest  number  of  rows  that  will 
keep  each  employed  an  exact  number  of  days  ? 

19.  A  grocer  has  a  4  pound,  5  pound,  6  pound,  and  a 
1 2  pound  weight :  what  is  the  smallest  tub  of  butter  that 
can  be  weighed  by  each  without  a  remainder  ? 


FRACTIONS. 

134.  A  Fraction,  is  one  or  more  of  the  equal  parts 
into  which  a  unit  is  divided. 

The  number  of  these  parts  indicates  their  name.  Thus, 
when  a  unit  is  divided  into  two  equal  parts,  the  parts  are 
called  halves;  when  into  three,  they  are  called  thirds;  when 
into  fou r,  fourths,  etc.  Hence, 

135.  A  half  is  one  of  the  two  equal  parts  of  a  unit ;  a 
third  is  one  of  the  three  equal  parts  of  a  unit ;  a  fourth, 
one  of  the  four  equal  parts,  etc. 

NOTE. — The  term  fraction  is  from  the  Latin  frango,  to  break. 
Hence,  Fractions  are  often  called  Broken  Numbers. 

136.  The  value  of  these  equal  parts  depends, 
First.  Upon  the  magnitude  of  the  unit  divided. 
Second.  The  number  of  parts  into  which  it  is  divided. 

ILLUSTRATION. — ist.  If  a  large  and  &  small  apple  are  each  divided 
into  two,  three,  four,  etc.,  equal  parts,  it  is  plain  that  the  parts  of  the 
former  will  be  larger  than  the  corresponding  parts  of  the  latter. 

2d.  If  one  of  two  equal  apples  is  divided  into  two  equal  parts,  and 
the  other  into  four,  the  parts  of  the  first  will  be  twice  as  large  as 
those  of  the  second ;  if  one  is  divided  into  two  equal  parts,  the  other 
into  six,  one  part  of  the  first  will  be  equal  to  three  of  the  second,  etc. 
Hence, 

NOTE. — A  half  is  twice  as  large  as  a  fourth,  three  times  as  large  as 
a  sixth,  four  times  as  large  as  an  eighth,  etc. ;  and  generally, 

The  greater  the  number  of  equal  parts  into  which  the  unit  is  di- 
vided, the  less  will  be  the  value  of  each  part.  Conversely, 

The  less  the  number  of  equal  parts,  the  greater  will  be  the  value 
of  each  part.' 

134.  What  is  a  fraction  ?  What  does  the  number  of  parts  indicate  ?  135.  What 
Is  a  half?  A  third?  A  fonrth?  A  tenth?  A.  hundredth  ?  136.  Upon  what  doei 
the  size  of  these  parts  depend  ?  99 


100  FRACTIONS. 

137.  Fractions  are  divided  into  two  classes,  common  and 
decimal. 

A  Common  Fraction  is  one  in  which  the  unit  is 
divided  into  any  number  of  equal  parts. 

138.  Common  fractions  are  expressed  by  figures  writ- 
ten above  and  below  a  line,  called  the  numerator   and 
denominator;  as  \,  f,  -j%. 

139.  The  Denominator  is  written  lelow  the  line, 
and  shows  into  how  many  equal  parts  the  unit  is  divided. 
It  is  so  called  because  it  names  the  parts;  as  halves,  thirds, 
fourths,  tenths,  etc. 

The  Numerator  is  written  above  the  line,  and 
shows  how  many  parts  are  expressed  by  the  fraction.  It 
is  so  called  because  it  numbers  the  parts  taken.  Thus,  iu 
the  fraction  f,  four  is  the  denominator,  and  shows  that 
the  unit  is  divided  into  four  equal  parts ;  three  is  the  nu- 
merator, and  shows  that  three  of  the  parts  are  taken. 

140.  The  Terms  of  a  fraction  are  the  numerator  and 
denominator^ 

NOTE. — Fractions  primarily  arise  from  dividing  a  single  unit  into 
equal  parts  They  are  also  used  to  express  a  part  of  a  collection  of 
units,  and  apart  of  &  fraction  itself;  as  i  half  of  6  pears,  i  third  of 
3  fourths,  etc.  .  But  that  from  which  they  arise,  is  always  regarded 
as  a  whole,  and  is  called  the  Unit  or  Base  of  the  fraction. 

141.  Common  fractions  are  usually  divided  into  proper, 
improper,  simple,  compound,  complex,  and  mixed  numbers. 

1.  A  Proper  Fraction  is  one  whose  numerator  is 
less  than-  the  denominator;  as  -£,  £,  f. 

2.  An  Improper  Fraction  is  one  whose  numerator 
equals  or  exceeds  the  denominator ;   as  £ ,  f . 

3.  A   Simple    Fraction   is    one    having  but  one 

137.  Into  how  many  classes  arc  fractions  divided?  A  common  fraction! 
> 38.  How  expressed  ?  m.  What  does  the  denominator  show  ?  Why  so  called? 
The  numerator  ?  Why  so  called  ?  140.  What  are  the  terms  of  a  fraction  ? 


FRACTION'S.  107 

numerator  and  one  denominator,  each  of  which  is  a  whole 
number,  and  may  be  proper  or  improper ;  as  f ,  |. 

4.  A  Compound  Fraction  is  a  fraction  of  a  frac- 
tion ;  as  |  of  f . 

5.  A  Complex  Fraction  is  one  which  has  a  frac' 

tional  numerator,  and  an  integral  denominator;  as,  -,  — . 

(  4    *> 

REMARK. — Those  abnormal  expressions,^&v\ngfnictionaljdenQm-^'' 

inators,  commonly  called  complex  fractions,  do  not  strictly  come  under 
the  definition  of  a  fraction.  For,  a  fraction  is  one  or  more  of  the 
equal  parts  into  which  a  unit  is  divided.  But  it  cannot  properly  be 
said,  that  a  unit  is  divided  into  s^ths,  4|dths,  etc. ;  that  is,  into  3^ 
equal  parts,  4!  equal  parts,  etc.  They  are  expressions  denoting  divi- 
sion, having  &  fractional  divisor,  and_should  be  treated  as  such. 

6.  A  Mixed  Nutnbe^\s>  a  whole  number  and  a  frac- 
tion expressed  together/fas,  5  f ,  344-j . 

NOTE. — The  primary  idea  of  a  fraction  is  a  part  of  a  unit.  Hence, 
a  fraction  of  less  value  than  a  unit,  is  called  a  proper  fraction. 

All  other  fractions  are  called  improper,  because,  being  equal  to  01 
greater  than  a  unit,  they  cannot  be  said  to  be  a  part  of  a  unit. 

Express  the  following  fractions  by  figures: 

1.  Two  fifths.  6.  Thirteen  twenty-firsts. 

2.  Four  sevenths.  7.  Fifteen  ninths. 

3.  Three  eighths.  8.  Twenty-three  tenths. 

4.  Five  twelfths.  9.  Thirty-one  forty-fifths. 

5.  Eleven  fifteenths.      10.  Sixty-nine  huudredths. 

11.  112  two  hundred  and  fourths. 

12.  256  five  hundred  and  twenty-seconds. 

13.  Explain  the  fraction  £. 

ANALYSIS. — \  denotes  \  of  i ;  that  is,  one  such  part  as  is  obtained 
by  dividing  a  unit  into  4  equal  parts.  The  denominator  names  the 
parts,  and  the  numerator  numbers  the  parts  taken.  It  is  common 
because  the  unit  is  divided  into  any  number  of  parts  taken  at  ran- 
dom ;  proper,  because  its  value  is  less  than  i  ;  and  simple,  because  it 
has  but  one  numerator  and  one  denominator,  each  of  which  is  a 
whole  number. 


141.  Into  what  are  common  fractions  divided  ?  A  proper  fraction  ?   Improper? 
Simple  ?    Compound  ?    Complex  ?    A  mixed  number  ? 


102  FEACTIONS. 

14.  Explain  the  fraction  £. 

ANALYSIS. — J  denotes  J  of  i,  or  \  of  3.  For,  if  3  equal  lines  are 
each  divided  into  4  equal  parts,  3  of  these  parts  will  be  equal  to  J  of 
i  line,  or  \  of  the  3  lines.  It  is  common,  etc.  (Ex.  13.) 

Read  and  explain  the  following  fractions : 

15.  f.  20.  4^-.  25.  1 6£. 

16.  $off.  21.  |f  26.  zSfJ. 

17.  -J-J-.  22.    ff.  27.    |  Of  |  Of  f 

1 8.  fofif  23.  If.  28.  I  off  of  2f 

ii  M_  IX^ 


NOTE. — The  2ist  and  22d  sliould  be  read,  "68  twenty  -first*," 
"  85  thirty  -seconds,"  and  not  68  twenty-on.es,  85  thirty-twos. 

142.  Fractions,  we  have  seen,  arise  from  division,  the 
numerator  being  the  dividend,  and  the  denominator  the 
divisor.     (Art.  64,  n.)     Hence, 

The  value  of  a  fraction  is  the  quotient  of  the  numerator 
divided  by  the  denominator.  Thus,  the  value  of  i  fourth 
is  i  ~-4f  or  £;  of  6  halves  is  6H-2,  or  3;  of  3  thirds  is 
3-^3>or  i. 

143.  To  find  a  Fractional  Part  of  a  number. 

i.  "What  is  i  half  of  12  dimes? 

ANALYSIS. — If  12  dimes  are  divided  into  2  equal  parts,  i  of  these 
parts  will  contain  6  dimes.  Therefore,  J ^  of  12  dimes  is  6  dimes. 

NOTE. — The  solution  of  this  and  similar  examples  is  an  illustra- 
tion of  the  second  object  or  office  of  Division.  (Art.  63,  b.) 

4.  What  is  ^  of  36  ?    |  of  45  ?     £  of  60  ?    |  of  63  ? 

5.  What  is  |  of  48  ?    £of63?    ^  of  190?    ^  of  132? 

6.  What  is  f  of  12  dimes? 

ANALYSIS. — 2  thirds  are  2  times  as  much  as  j.  But  i  third  of  12 
is  4.  Therefore,  §  of  12  dimes  are  2  times  4,  or  8  dimes.  Hence,  the 

142.  From  what  do  fractions  arise  ?  Which  part  is  the  dividend  ?  Which  the 
ot'isor  ?  What  is  the  value  of  a  fraction  > 


FRACTIONS.  103 

RULE. — Divide  the  given  number  by  the  denominator,  and 
multiply  the  quotient  by  the  numerator 

To  find  £,  divide  the  number  by  2. 
To  find  |,  divide  the  number  by  3. 
To  find  f,  divide  by  3,  and  multiply  by  2,  etc. 

7.  What  is  |  of  56  ?  Ans. 

8.  What  is  f.  of  30  apples  ?     Of  45  ?     Of  60  ? 
^9.  What  is  £  of  42  ?    }  of  56  ?    f  of  72  ? 

foTWhat  is  f  of  45  ?    -^  of  50  ?    ^  of  .66-?^T77  of  108  ? 


GENERAL  PRINCIPLES   OF   FRACTIONS. 

144.  Since  the  numerator  and  denominator  have  the 
tame  relation  to  each  other  as  the  dividend  and  divisor,  it 
follows  that  the  general  principles  established  in  division 
are  applicable  to  fractions.  (Art.  81.)  That  is, 

1.  If  the  numerator  is  equal  to  the  denominator,  the 
value  of  the  fraction  is  i. 

2.  If  the  denominator  is  greater  than  the  numerator,  the 
value  is  less  than  i. 

3.  If  the  denominator  is  less  than  the  numerator,  the 
ralue  is  greater  than  i. 

4.  If  the  denominator  is  i,  the  value  is  the  numerator. 

5.  Multiplying  the  numerator,  multiplies  the  fraction. 

6.  Dividing  the  numerator,  divides  the  fraction. 

—  7.  Multiplying  the  denominator,  divides  the  fraction. 

8.  Dividing  the  denominator,  multiplies  the  fraction. 

9.  Multiplying  or  dividing  both  the  numerator  and  de- 
nominator by  the  same  number  does  not  alter  the  value  of 
the  fraction. 


144.  If  the  numerator  Is  equal  to  the  denominator,  what  Is  the  value  of  th« 
fraction  ?  If  the  denominator  Is  the  greater,  what  ?  If  less,  what  ?  If  the  de- 
nominator Is  i,  what  ?  What  Is  the  effect  of  multiplying  the  numerator  ?  Divid- 
ing the  numerator  f  Multiplying  the  denominator  ?  Dividing  the  denominator  ? 


104  FRACTIONS. 


REDUCTION    OF    FRACTIONS. 

145.  Reduction  of  fractions  is  changing  their  terms, 
without  altering  the  value  of  the  fractions.     (Art.  142.) 

CASE    I. 

146.  To  Reduce  a  Fraction  to  its  Lowest  Terms. 

DBF. — The  Lowest  Terms  of  a  fraction  are  the 
smallest  numbers  in  which  its  numerator  and  denominator 
can  be  expressed.  (Art.  101,  Def.  7.) 

Ex.  i.  Reduce  ff  to  its  lowest  terms. 

ANALYSIS. — Dividing  both  terms  of  a  fraction  ist  METHOD. 

by  the  same  number  does  not  alter  its  value.  ,24       12 

(Art.  144,  Prin.  9.)      Hence,  we  divide  the  given  2' ^2       7(5  °~ 

terms  by  2,  and  the  terms  of  the  new  fraction  I2       ? 

by  4  ;  the  result  is  1.    But  the  terms  of  the  frac-  4)~~7  ==  ~     Ans. 
tion  J  are  prime  to  each  other ;  therefore,  they 
are  the  lowest  terms  in  which  f -J  can  be  expressed. 

Or,  if  we  divide  both  terms  by  their  greatest          *L  XETHOD. 
common  divisor,  which  is  8,  we  shall  obtain  the         24     3 
same  result.     (Art.  126.)     Hence,  the  3^=4          ' 

RULE. — Divide  the  numerator  and  denominator  con- 
tinually by  any  number  that  will  divide  both  without  a  re- 
mainder, until  no  number  greater  than  i  will  divide  them. 

Or,  divide  both  terms  of  the  fraction  by  their  greatest 
common  divisor.  (Art.  126.) 

NOTES. — i.  It  follows,  conversely,  that  a  fraction  is  reduced  to 
higher  terms,  by  multiplying  the  numerator  and  denominator  by  a 
common  multiplier.  Thus,  §=H>  both  terms  being  multiplied  by  8. 

2.  These  rules  depend  upon  the  principle,  that  dividing  both  terms 
of  a  fraction  by  the  same  number  does  not  alter  its  value.  When  the 
terms  are  large,  the  second  method  is  preferable. 


145.  What  is  Reduction  of  Fractions*    146.  What  are  the  lowest  terms  of  a 
fraction  •    How  reduce  a  fraction  to  its  lowest  terms  f 


SEDUCTION     OF     FRACTIONS.  105 

Eeduce  the  following  fractions  to  their  lowest  terms : 

2.  ||.  9.  $f  1 6.  iff.  23. 

3-  M-  I0-  iH-  J7-  fj-  24. 

4-  if-  ".  %H.  1 8.  f^f  25.  |f§. 
5.  ff.  12.  f|f.  19.  ,flfr.  26.  #ft. 

6-  H-  13-  -Hi-  20.  #$.  27.  Htf. 

7-  ft-  14-    T7^-  21.    -fl&.  28.    i^. 

8-  &  15-    if-  22.    -#&.  29.    -ftV 

CASE    II. 

147.    To    reduce    an    Improper    Fraction   to    a    Whole   or 
Mixed  Number. 

i.  Eeduce  ^5-  to  a  whole  or  mixed  number. 

ANALYSIS. — Since  4  fourths  make  a  unit  or  i,  45      4)45 
fourths  will  make  as  many  units  as  4  is  contained  T    A  „„ 

times  in  45,  which  is  n^.    Hence,  the 

ETJLE. — Divide  the  numerator  ~by  the  denominator. 

NOTES. — i.  This  rule,  in  effect,  divides  both  terms  of  the  fraction 
by  the  same  number ;  for,  removing  the  denominator  cancels  it,  and 
cancelling  the  denominator  divides  it  by  itself.  (Art.  121.) 

2.  If  fractions  occur  in  the  answer,  they  should  be  reduced  to  the 
lowest  terms. 

Eeduce  the  following  to  whole  or  mixed  numbers : 
2.  i-Jl.  8.  -ip-.  14.  -^V--  20.  ± 

3-  *$*••  9-  -4i423"-  J5-  "Wr- 

4.    1 1 -2-.  10.   -§^-.  l6.    Tggg°.  22.   -3J 

5-  ¥/-•          ii.  W-          17-  -ViW          23.  ^ 

6.  -!•££.  12.  ^8/.  1 8.  -^fl^.  24 

7-  W^  J3-  Hf«  J9-  Hfl^-  25, 

26.  Iix  ^f fir5-  of  a  pound,  how  many  pounds  ? 

27.  In  VzVW-  °f  a  dollar,  how  many  dollars  ? 

28.  In  25-^p^-  of  ?.  y^ar,  how  many  years? 


147,  Bow  reduce  an  improper  fraction  to  a  whole  or  mixed  number  T 


106  DEDUCTION     OF     FRACTIONS. 

CASE    III. 

148.  To  reduce  a  Mixed  Number  to  an  Improper  Fraction. 

i.  Reduce  8f  to  an  improper  fraction. 

ANALYSIS. — Since  there  are  7  sevenths  in  a  unit,  there      ga 
must  be  7  times  as  many  sevenths  in  a  number  as  there       _ 
are  units,  and  7  times  8  are  56  and  3  sevenths  make  ty.      ^— 
Therefore,  8*=^.  *f  Am. 

Or  thus:  Since  i=$,  8  =  8  times  \  or  &P-,  and  *  make  V.  In  the 
operation  we  multiply  the  integer  by  the  given  denominator,  and  to 
the  product  add  the  numerator.  Hence,  the 

RULE. — Multiply  the  whole  number  by  the  given  denomi- 
nator;  to  the  product  add  the  numerator,  and  place  the 
sum  over  the  denominator. 

REMAKK. — A  whole  number  may  be  reduced  to  an  improper 
fraction  by  making  I  its  denominator.  Thus,  4=f .  (Art.  81.) 

Reduce  the  following  to  improper  fractions : 

2.  isf  6.  i45ff  10.  i573f  14- 

3.  i8f  7-  i»7H-  "•  256if  15- 
4-  35f  8-  295i¥ff-  12.  364of.  1 6. 
5.  81^.  9.  806^.  13.  8624^-  17- 

1 8.  In  263-^5-  pounds  how  many  sixteenths  ? 

19.  Change  641^  mile  to  fortieths  of  a  mile. 

CASE    IV. 

149.  To  reduce  a  Compound  Fraction  to  a  Simple  one. 

i.  Reduce  |  of  $  to  a  simple  fraction. 

ANALYSIS. — 3  fourths  of  f  =  3  times  i  fourth  of  f.  iBt  METHOD. 

Now  i  fourth  of  J  is   ft  ;  for  multiplying  the  de-  326 

nominator  divides   the   fraction:   and  3   fourths   of  4X3~~72 
|  =  3  times  -ft  or  &  ;  for  multiplying  the  numerator 

multiplies  the  fraction.     (Art.  144.)    Reduced  to  its  A             -_* 

lowest  terms  i62=£.  '  12      2 


148.  How  reduce  a  mixed  number  fQ  an  Improper  fraction  T    Bern,  A  wholq 
pnnrt>er  to  an  Improper  ft-action  T 


REDUCTION     OF     FRACTIONS.  107 

Or,    since    numerators    are    dividends    and  *d  METHOD. 

denominators    divisors,  the    factors  2   and   3,  i,  $     2      i 

which  are  common  to  both,  may  be  cancelled.  2  7  x  «~~  Ans. 
(Art.  144,  Prin.  9.)    The  result  is  £.    Heuce,  the 

RULE.  —  Cancel  the  common  factors,  and  place  the  product 
of  the  factors  remaining  in  the  numerators  over  the  product 
of  those  remaining  in  the  denominators. 

NOTES.  —  i.  Whole  and  mixed  numbers  must  be  reduced  to  improper 
fractions,  before  multiplying  the  terms,  or  cancelling  the  factors. 

2.  Cancelling  the  common  factors  reduces  the  result  to  the  lowest 
terms,  and  therefore  should  be  employed,  whenever  practicable. 

3.  The  numerators  being  dividends,  may  be  placed  on  the  right 
of  a  perpendicular  line,  and  the  denominators  on  the  left,  if  pre- 
ferred.   (Art.  122,  Bern.) 

4.  The  reason  of  the  rule  is  this:  multiplying  the  numerator  of 
one  fraction  by  the  numerator  of  another,  multiplies  the  value  of  the 
former  fraction  by  as  many  units  as  are  contained  in  the  numerator 
of  the  latter  ;  consequently  the  result  is  as  many  times  too  large  as 
there  are  units  in  the  denominator  of  the  latter.     (Art.  144,  Prin.  5.) 
This  error  is  corrected  by  multiplying  the  two  denominators  to- 
gether.    (Art.  144,  Prin.  7.) 


2.  Reduce  f  of  £  of  ~fa  of  2  J  to  a  simple  fraction.   Ans.  $£. 
Reduce  the  following  to  simple  fractions: 

S-fof-ft-  9.  I  of  A  of  60.  15-  A  of  ft  of  ft. 

4.  i  of  f  .  10.  |  of  H  of  ffr.  1  6.  |  of  f  of  |  of  f 

5.  4  of  ^.  ii.  &  of  H  of  IJ.  17.  H  of  1J  of  65}. 

6.  |  of  -H.  12.  Jf  of  |  of  |f.  1  8.  fj-  of  ff  of  84J 

7.  ^  of  «.  13.  ||  of  A  of  45-  19-  I  of  8f  of 

8.  f  of  tV.  14.  i  of  f  of  ||.  20.  £  of  |  of 


21.  Reduce  £|  of  4$  of  £$  of  9!  to  a  simple  frac- 
tion. 

22.  To  what  is  |  of  fj  of  3^  bushels  equal? 

23.  To  what  is  f  of  4^  of  -fo  of  a  yard  equal  ? 


149.  How  reduce  a  compound  fraction  to  a  simple  one  T    Note.  What  must  be 
done  with  wbole  aod  mjjed  oumberi  T    What  i»  the  advantage  of  cancelling  r 


108  BEDUCTION     OF     FEACTIONS. 

CASE     V. 
150.  To  reduce  a  Fraction  to  any  required  Denominator. 

1.  Keduce  f  to  twenty-fourths. 

IST  ANALYSIS.  —  8  is  contained  in  24,  3  times  ;  there-  24-1-8:=  3 
fore,  multiplying  both  terms  by  3,  the  fraction  be-  6x3  18 
comes  if,  and  its  value  is  not  altered.  (Art.  144.)  8  X  V  =  ~ 

REM.  —  If  required  to  reduce  £  to  fourths,  we  should  divide  both 
terms  by  2,  and  it  becomes  J.  (Art.  144.) 

20  ANALYSIS.  —  Multiplying    both    terms  of   f  6x24=144 

by  24,  the    required    denominator,  we    have  |Jf.  g  ^  ,  .  _  ~7T 
(Art.  144,  Prin.  9.)    Again,  dividing  both  terms  of  •}  -»  * 

by  8,  the  given  denominator,  we  have  i£,  the  same  — 

as  before.    Hence,  the  192-7-8     124 

EULE.  —  Multiply  or  divide  loth  terms  of  the  fraction 
by  such  a  number  as  will  make  the  given  denominator 
equal  to  the  required  denominator. 

Or,  Multiply  both  terms  of  the  given  fraction  by  the 
required  denominator  ;  then  divide  both  terms  of  the  re- 
sult by  the  given  denominator.  (Art.  144,  Prin.  9.) 

NOTES.  —  i.  The  multiplier  required  by  the  ist  method  is  found 
by  dividing  the  proposed  denominator  by  the  given  denominator. 

2.  When  the  required  denominator  is  neither  a  multiple,  nor  an 
exact  divisor  of  the  given  denominator,  the  result  will  be  a  complex 

3V  4* 

fraction.     Thus,  ^  reduced  to  tenths=—  ;  £  reduced  to  sevenths=  — 

Change  the  following  to  the  denominator  indicated  : 

2.  -^  to  fifty-seconds.  7.  f  £  to  246ths. 

3.  -fa  to  sixtieths.  8.  %\  to  288ths. 

4.  -jUj-  to  eighty-eighths.  9.  f&  to  36oths. 

5.  -££  to  i44ths.  10.  T^  to  loooths. 

6.  to  2O4ths.  ii.  ^°  looooths. 


12.  Keduce  f  to  fourths.  Ans.  —. 

1  3.  Reduce  ^  to  sixths.    15.  Reduce  f  to  twenty-sevenths. 
14.  Reduce  £  to  thirds.    16.  Reduce  -f£  to  fourths. 


ijo.  How  reduce  a  fraction  to  any  required  den.pmin»ticm  ? 


REDUCTION     OF     FRACTIONS.  109 

151.  To  reduce  a  Whole  Number  to  a  Fraction  having  a 
given  Denominator. 

1.  Eeduce  7  to  fifths. 

ANALYSIS. — ist.  Since  there  are  5  fifths  in  a  unit,  an>  number 
must  contain  5  times  as  many  fifths  as  units ;  and  5  time&  7  are  35. 
Therefore  7=a£. 

Or,  2d.  Since  i=f,  7  must  equal  7  times  £, 
or  3sa.    In  the  operation  7  is  both  multiplied  OFBKATIOH. 

and  divided  by  5;  hence,  its  value  is  not    7=^(7  x  5) -1-5— -5*. 
altered.    (Art.  86,  a.)    Hence,  the 

RULE. — Multiply  the  whole  number  by  the  given  denomi* 
nator,  and  place  the  product  over  it. 

2.  Change  7  to  a  fraction  having  9  for  its  denominator. 

3.  Change  63  to  5ths.  9.  468  to  76ths. 

4.  Reduce  79  to  7ths.  10.  500  to  87ths. 

5.  Reduce  83  to  gths.  n.  1560  to  hundredths. 

6.  Reduce  105  to  i6ths.  12.  2004  to  thousandths. 

7.  Reduce  217  to  2oths.  13.  500  to  ten-thousandths. 

8.  Reduce  321  to  49ths.  14.  25  to  millionths. 

CASE   VI. 
152.  To  reduce  a  Complex  Fraction  to  a  Simple  One, 

i.  Reduce  the  complex  fraction  —  to  a  simple  one. 

0 

ANALYSIS. — The  denominator  of  a  fraction,  we  OPBKATIOH. 
have  seen,  is  a  divisor.     Hence,  the  given  complex  3! 
fraction  is  equivalent  to  3^-2-5.     (Art.  142.)    Now  ~  — 3T~^~5> 
3i=J£;  therefore,  3i-«-5=¥-5-5.    (Art.  148.)    But  ?  ,l=ip  . 
•^^-5= $ ;  for,  dividing  the  numerator  by  any  num- 
ber, divides  the   fraction  by  that  number.     (Art.  '7~r5~3r- 
144,  Prin.  6.)     Therefore,  f  is  the  simple  fraction  Ans.  f. 
required. 

Or,  multiplying  the  denominator  by  5,  divides  the  fraction,  and 
we  have  !£=$,  the  same  as  before.  (Prin.  7.) 

150.  Upon  what  principle  IB  this  mle  founded  ?  151.  How  redoce  ft  wholp 
number  to  &  fraction  having  a  given  denominator  ? 


'10  RRDUCTIOK     OP     FRACTIONS. 

A 

2.  Reduce  the  complex  fraction  -  to  a  simple  one. 

ANALYSIS.'— Reasoning  as  before,  the  given  fraction        OPIBJ  non. 

ts  equivalent  to  f -r-y.      But  we  cannot  divide  the  4 

numerator  of  the  fraction  4  by  7  without  a  remainder,  ~  =  i~^~  7  5 
we  therefore  multiply  the  denominator  by  it ;   for, 

multiplying  the  denominator,  divides  the  fraction;  *~7  —  Ts- 
and  ^-T-7=2\,  the  simple  fraction  required.    (Art.  144,        Ans.  ^8. 
Prin.  7.)    Hence,  the 

KULE. — I.  Reduce  the  numerator  of  the  complex  fraction 
to  a  simple  one. 

II.  Divide  its  numerator  by  the  given  denominator. 
Or,  Multiply  its  denominator  by  the  given  denominator. 

NOTES. — i.  When  the  numerator  can  be  divided  without  a  remain- 
the  former  method  is  preferable.  When  this  cannot  be  dona 
tbe  latter  should  be  employed. 

s.  After  the  numerator  of  the  complex  fraction  is  reduced  to  a 
simple  one,  the  factors  common  to  it  and  the  given  denominator, 
should  be  cancelled. 


If  this  case  is  deemed  too  difficult  for  beginners,  it  may  be 
fitted  till  review. 

(For  the  method  of  treating  those  expressions  which  have  fractional 
•'mominatora,  see  Division  of  Fractions,  p.  130.) 

c-s 
2.  Reduce  ~  to  a  simple  fraction. 

24 
SOLUTION. — 5$=^  ;  and  ^-H  24 =->*&,  or  ££,  Ant. 

Reduce  the  following  complex  fractions  to  simple  ones. 

,  f . 


8.  -Z-.  12. 


6.  ^-.  10. 

2  2 


152.  How  reduce  a  complex  fraction  to  a  simple  one?    Not*.  What  should  bq 
done  with  common  factors  ? 


BEDUCTION     OF     FBACTIONS.  Ill 


CASE     VII. 
153.  To  reduce  Fractions  to  a  Common  Denominator. 

DBF.  —  A  Common  Denominator  is  one  that 
belongs  equally  to  two  or  more  fractions  ;  as,  f,  f,  f. 

i.  Keduce  ^,  f  ,  f  to  a  common  denominator. 

REMARK.  —  In  reducing  fractions  to  a  common  denominator,  it 
•hould  be  observed  that  the  value  of  the  fractions  is  not  to  be  altered. 
Hence,  whatever  change  is  made  in  any  denominator,  a  corresponding 
change  must  be  made  in  its  numerator. 

ANALYSIS.  —  If  each  denominator  is  mul-  OPERATION. 

tiplied  by  all  the  other  denominators,  the  i        1x3x4        12 

fractions  will  have  a  common  denominator.  ~  =  2  X.  1  X  d.  ==  24* 
and  if  each  numerator  is  multiplied  into  all 

the  denominators  except  its  own,  the  terms  £  __  2  x  2  x  4  _  £J 

of  each  fraction  will  be  multiplied  by  the  3        3x2x4        24 

same  numbers  ;   consequently  their  value  ,        7  x  2  x  ^        18 

will  not  be  altered.    (Art.  144.)    The  frac-  -  =  —     —  -  =  —  . 

tions  thus  obtained  are  iJ,  if,    and    if.  4       4x2x3        24 
Hence,  the 

RULE.  —  Multiply  the  terms  of  each  fraction  by  all  the 
denominators  except  its  own. 

NOTES.  —  i.  Mixed  numbers  must  first  be  reduced  to  improper 
fractions,  compound  and  complex  fractions  to  simple  ones. 

2;  This  rule  is  founded  upon  the  principle,  that  if  both  terms  of  a 
fraction  are  multiplied  by  the  same  numbers,  its  value  is  not  altered. 

3.  A  common  denominator,  it  will  be  seen,  is  the  product  of  all  the 
denominators  ;  hence  it  is  a  common  multiple  of  them.  (Art.  129.) 

Reduce  the  following  to  a  common  denominator  : 

2.  |  and  f  5.  $,  |,  and  -ft.  8.  f,  f,  tf,  #. 

3.  i,  |,  and  f  .        6.  -f,  T6T,  and  A-          9-  H>  ft, 

4.  |,  f  ,  and  -A,      7-  I  i,  A»  and  ^      10.  -fi  ||, 


153.  What  la  a  common  denominator?  How  reduce  fractions  to  a  common 
denominator  ?  Note.  What  is  this  rnle  based  upon  ?  What  ia  to  be  done  wlttj 
mixed  numbers,  compound  and  complex  fractions  ? 


112  REDUCTION     OF     FRACTIONS. 

ii.  Find  a  common  denominator  of  4,  3^,  and  £  of  | 

ANALYSIS.— 4  =  } ;  3$  =  £  ;  and  $  of  £  =  \.    Reducing  t,  3    fe 
to  a  common  denominator,  they  become  38%  ^  and  f . 

12.  Keduce  3^,  i|,  2-$-.  15.  Reduce  5-$-,  \  of  8,  •{£. 

13.  Reduce  6£,  7-fJ,  -,%-,  |.      16.  Reduce  f  of  \  of  9,  i  i-J,  /p 

14.  Reduce  f  of  },  5$-,  f.       17.  Reduce  13^,  17,  f  of  f. 

1 8.  Find  a  common  denominator  of  2^,  5-^  and  2f. 

19.  Reduce  -^,  f,  and  ££,  to  the  common  denominator 
100. 

SOLUTIOK.— The  fraction  -&=-&&',  £=-Mr;  and  M=\t£a- 

20.  Change  |,  f,  and  f  to  72ds. 
^I.  Change  f,  T7^,  and  {£  to  96ths. 

22.  Reduce  f  of  f  and  •££  to  poths. 

23.  Reduce  ^  of  lof  and  ^,  to  75ths. 

24.  Reduce  f,  -j^,  and  ^|  to  i68ths. 

25.  Reduce  -^5-,  f§,  and  fj$,  to  icooths. 


CASE   VIII. 

154.    To    Reduce    Fractions  to   the  Least   Common 
Denominator. 

DEF. — A  Common  Denominator  is  a  common 
multiple  of  all  the  denominators.     (Art.  153,  n.)     Hence, 

The  Least  Common  Denominator  is  the  least 
common  multiple  of  all  the  denominators. 

i.  Reduce  f,  ^,  and  -f|  to  the  least  common  denomi- 
nator. 

ANALYSIS. — Here   are  two  steps :    1st.  To  find  OPERATION. 

the  least  common  multiple  of  the  denominators ;  3)3>  I2>  J5 

2d.  To  reduce  the  given  fractions  to  this  denomi-  ~^      T      r 

nator.    The  least  common  multiple  of  the  given  5  x  4  x  C  —  60 
denominators  is  60.     (Art.  133.) 

154.  What  is  the  least  common  denominator?    How  reduce  fractions  to  the 
least  common  denominator  ?    Note.  What  is  to  be  done  with  mixed  numbers, 
>und  and  complex  fractions  ? 


REDUCTION     OF    .FRACTIONS.  113 

To  reduce  the  given  fractions  to  6oths,  we  multiply  both  terms  of 
|  by  20,  and  it  becomes  $g.  In  like  manner,  if  both  terms  of  T6^  are 
multiplied  by  5,  it  becomes  ££  ;  and  multiplying  both  terms  of  |£by 
4,  it  becomes  $$.  Therefore,  f ,  ^f,  and  {%  are  equal  to  $g,  f  $  and 
|^.  Hence,  the 

RULE. — Find  the  least  common  multiple  of  all  the  denom- 
inators, and  multiply  both  terms  of  each  fraction  by  such  a 
number  as  will  reduce  it  to  this  denominator.  (Art.  150,  n.} 

NOTES. — i.  Mixed  numbers  must  be  reduced  to  improper  frac- 
tions ;  compound  and  complex  fractions  to  simple  ones ;  and  all  frac- 
tions to  the  lowest  terms,  before  applying  the  rule.  If  not  reduced 
to  the  lowest  terms,  the  least  common  multiple  of  the  denominators 
is  liable  not  to  be  the  least  common  denominator.  (See  Ex.  2,  18.) 

2.  This  rule,  like  the  preceding,  is  based  upon  the  principle  that 
multiplying  both  terms  of  a  fraction  by  the  same  number  does  not 
alter  its  value.  (Art.  144,  Prin.  9.) 

2.  Reduce  15,  2^,  f  of  f,  and  •&  ^°  the  least  common 
denominator. 

15  7    2    ,$       2       ,  6        i 

ANALYSIS. — 1 1  =  — -,  2-J-  =  -.  -  of  -  =  -,  and  —  —  - 

i  3*55          12       2 

Now    the    least     common    multiple    of    the    denominators    of 

15    7    2    I    .  .        450    70    12    15 

—    -    -   -,  is  30.       A.ns. —    —    — . 

I     3    5    2  30     30    30    30 

Reduce  the  following  to  least  common  denominator : 

?      2     3      5  TT        8        6.     14     C2 

6"    T>  3>  7'  •••   TfTS>  TT5>   55>  J^' 

6.  y,  £,  4^.  14.  ^  of  13, 

8.  A>  1?  f  of  to.  16. 

10.  ^  of  8|,  i  of  40.        18.  ifj,  $ffi,  H||. 


114  ADDITION    OF    FBACTIONS. 


ADDITION   OF   FRACTIONS. 

155.  Addition  of  Fractions  embraces  two  classes 
of  examples,  viz. :  those  which  have  a  common  denominator 
and  those  which  have  different  denominators. 

156.  When  two  or  more  fractions  have  a  common  de- 
nominator, and  refer  to  the  same  kind  of  unit  or  base, 
their  numerators  are  like  parts  of  that  unit  or  base,  and 
therefore  are  like  numbers.    Hence,  they  may  be  added, 
subtracted,  and  divided  in  the  same  manner  as  whole  num- 
bers.   Thus,  f  and  •£  are  12  eighths,  just  as  5  yards  and 
7  yards  are  12  yards.    (Art.  101,  Def.  13.) 

157.  "When  two  or  more  fractions  have  different  denom- 
inators, their  numerators  are  unlike  parts,  and  therefore 
cannot  be  added  to  or  subtracted  from  each  other  directly, 
any  more  than  yards  and  dollars.     (Art.  101,  Def.  14.) 

158.   To  add   Fractions  which   have  a   Common  Denom- 
inator. 

1.  What  is  the  sum  of  f  yard,  f  yard,  and  I  yard  ? 

ANALYSIS. — 3  eighths  and  5  eighths  OPERATION. 

yard  are  8  eighths,  and  7  are  15  eighths,     f  +  f  +  i=  V ,  or  l|  y. 
<  qual  to  I J  yard.    (Art.  147.)    For,  since 

the  given  fractions  have  a  common  denominator,  their  numerators 
are  like  numbers.     Hence,  the 

KFLE. — Add  the  numerators,  and  place  the  sum  over  the 
common  denominator. 

NOTE. — The  answers  should  be  reduced  to  the  lowest  terms ;  and  If 
'improper  fractions,  to  whale  or  mixed  numbers. 

Add  the  following  fractions . 

2.  f,  f,  and  f  5.  -rtfr,  -flfr,  -£&,  and  ft%. 

3-  A,  A>  A,  and  ft.  6.  &,  rfft,  «|,  and 

4-  A.  A,  A,  and  H-  7-  m>  -Hi,  m  and 

158.  How  a<W  fractions  that  have  a  common  denominator  T 


ADDITION     OF     FRACTIONS.  115 

159.  To  add  Fractions  which  have  Different  Denomina- 

tors. 

8.  Find  the  sum  of  £  of  a  pound,  f  of  a  pound,  and  f  of 
a  pound. 

ANALYSIS.  —  As  these  fractions  OPBKATION. 

have     different     denominators,  4  x  5  x  8=160,  com.  d. 

their  numerators  cannot  be  added  1x5x8=   40,  1st  nu. 

in  their  present  form.    (Art.  157.)  2x4x8=:    64,  2d     " 

We  therefore  reduce  them  to  a  5x4x5  —  100,  3d     " 

common  denominator,  which  is  Sum  of  nu.,  204.     Hence 

160.  Then  i=tfft,  i=-ftfc,  and  4<D       64       100      204 

'  =  |§§.    Adding  the  numerators,     —7-  +  ~7~  +  ~  —  —  ~  7~;  or  l£-£ 

loo     loo     160     160 
and  placing  the   sum  over   the 

common  denominator,  we  have  f$£,  or  i^,,  the  answer  required. 
For,  reducing  fractions  to  a  common  denominator  does  not  alter 
their  value  ;  and  when  reduced  to  a  common  denominator,  the  nu- 
merators are  like  numbers.  (Arts.  153,  156.) 

Or,  we  may  reduce  them  to  the  least  common  denominator,  which 
is  40,  and  then  add  the  numerators.  (Art.  154.)  Hence,  the 

EULE.  —  Reduce  the  fractions  to  a  common  denominator, 
and  place  the  sum  of  the  numerators  over  it. 

Or,  reduce  the  fractions  to  the  least  common  denomina- 
tor, and  over  this  place  the  sum  of  the  numerators. 

REMARKS.  —  i.  The  fractional  and  integral  parts  of  mixed  num- 
bers should  be  added  separately,  and  the  results  be  united. 

Or,  whole  and  mixed  numbers  may  be  reduced  to  improper  frac- 
tion*, then  be  added  by  the  rule.  (Art.  148.) 

2.  Compound  and  complex  fractions  must  be  reduced  to  simple 
ones  ;  then  proceed  according  to  the  rule.  (Ex.  20,  28.) 

(For  Addition  of  Denominate  Fractions,  see  Art.  315.) 

9.  What  is  the  sum  of  5,  2  J,  and  io|  ? 

ANALYSIS.  —  Reducing  these  fractions  to  a  common  5    =e 

denominator  12,  we  have,  5  =  5  ;  2^=2,\;  and  ioj=  2l  =  2  4 
ioyV,      Adding  the  numerators.  4  twelfths    and  9          3  o 

twelfthsare  if  —  i-j\.    i  and  10  are  n  and  2  are  13  and  ___  *~      " 


5  are  18.     Ans.  i8rV-    Or,  s=f  or  ft;  2i=f$  ;  and      Ans.  18-^ 
iol=\^.    Now  f  $  +  f  I  +  -L,¥=  W,  or  i8-,V  Ans. 

159.  How  when  they  have  not  T    Note.  How  add  whole  and  mixed  numbers  ? 
Compound  and  complex  fractious*  f 


116                    ADDITION  OF     FRACTION'S. 

Add  the  following: 

(10.)          (ii.)  (12.)          (13.) 

'*             4*  27-A            «             ft 

4              Si  «A            tt             H 


(iS.)  (16.)  (17.)  (18.)  (19.) 

19^  244  207!  175!  450 

47i  i°H-  62|  207  67^ 

68$  68^  49^  368^ 


20.  What  is  the  sum  of  £  of  f  ,  J-  of  7,  and  £  of  pf  ? 

21.  What  is  the  sum  of  f  of  -f-,  |-  of  f,  and  f  of  7  ? 

22.  What  is  the  sum  of  f  of  4^,  f  of  3,  and  -£  of  10^  ? 

23.  A  beggar  received  $if  from  one  person,  $2^  from 
another,  and  $3!  from  another  :  how  much  did  he  receive 
from  all  ? 

24.  If  a  man  lays  up  $43!  a  month,  and  his  son  $27^, 
how  much  will  both  save  ? 

25.  What  is  the  sum  of  f  ,  f  ,  £J,  and  f  pound  ? 

26.  A  shopkeeper  sold  17^  yards  of  muslin  to  one  cus- 
tomer, 8^  yards  to  another,  25^  to  another  :  how  many 
yards  did  he  sell  to  all  ? 

27.  A  farmer  paid  $i8f  for  hay,  $45-fV  for  a  cow,  $150! 
for  a  horse,  and  $275  for  a  buggy:  how  much  did  he  give 

for  all? 

.1    fi  a 

28.  What  is  the  sum  of  2f,  —  ,  and  ^.  , 

53  4 

ANALYSIS.  —  Re.  lucing  the  complex  fractions  to  simple  ones,  we 

have,  **=l+s=-?e  ;  -5i=V+3=*  5  and  -=3-*-4=A,  or  i     Now, 
53  4 


.,.  Add  ^  «*,  and  2Jt    30.  Add  U,  f£,  and  3. 

23  7  573 


SUBTRACTION     OF     FKACTIONS.  117 


SUBTRACTION    OF    FRACTIONS. 

160.  Subtraction  of  Fractions  embraces  tiao 
classes  of  examples,  viz.:   those  which   have  a  common 
denominator,  and  those  which  have  different  denominators. 

161.  To   subtract    Fractions   which    have   a    Common 

Denominator. 

1.  What  is  the  difference  between  ||  and  ^f  ? 
ANALYSIS.  —  13  sixteenths  from  15  sixteenths 

.,          x,  •       j        T-I  OPERATION. 

leave  2  sixteenths,  the   answer  required.     For, 

since  the  given  fractions  have  a  common  denomi-  _£  __  3  __  _ 

nator,  their  numerators,  we  have  seen,  are  like  JO        1  6        1  6 
numbers.    (Art.  156.)    Hence,  the 

RULE.  —  Take  the  less  numerator  from  the  greater,  and 
place  the  difference  over  the  common  denominator. 

2.  From  f$  take  ft.  5.  From  fi|  take  iff 

3.  From  ff  take  $f.  6.  From  f$|  take  f°-|. 

4.  From  m  take  ^V  7-  From  T^&  take  ^ 


162.    To   subtract   Fractions  which   have  Different 
Denominators. 

8.  From  £  of  a  pound,  subtract  |  of  a  pound.  6  24 

ANALYSIS.—  Since    these    fractions    have    different  j  28 

denominators,  their  numerators  cannot  be  subtracted  3  21 

in  their  present  form.     We  therefore  reduce  them  to  a  "7  =  To 
common  denominator,  which  is  28  ;  then  subtracting  as 

above,   have    £g,  the    answer    required.      (Art.    153.)  Ans  — 

Hence,  the  28 

RULE.  —  Reduce  the  fractions  to  a  common  denominator, 
and  over  it  place  the  difference  of  the  numerators. 

161.  How  subtract  fractions  that  have  a  common  denominator?  162.  How 
•when  they  have  different  denominators  ?  Hem.  How  subtract  mixed  numbered 
Compound  and  complex  fractions  ?  A  proper  fraction  from  a  whole  number  T 


118  8UBTBACT10K    OF     FRACTIONS. 

REMARKS. — i.  The  fractional  and  integral  parts  of  mixed  numbers 
should  be  subtracted  separately,  and  the  results  be  united. 

Or,  they  may  be  reduced  to  improper  fractions,  then  apply  the  rule. 

Compound  and  complex  fractions  should  be  reduced  to  simple  ones, 
and  all  fractions  to  their  lowest  terms. 

2.  A.  proper  fraction  may  be  subtracted  from  a  wJtole  number  by 
taking  it  from  a  unit;   then  annex  the  remainder  to  the  whole 
number  minus  1. 

Or,  the  whole  number  may  be  reduced  to  a  fraction  of  the  same 
denominator  as  that  of  the  given  fraction  ;  then  subtract  according 
to  the  rule.  (Art.  151.) 

3.  The  operation  may  often  be  shortened  by  finding  the  least  com 
mon  denominator  of  the  given  fractions.     (Art.  154.) 

(For  subtraction  of  Denominate  Fractions,  see  Art.  317.) 

9.  What  is  the  difference  between  1 2\  pounds  and  5$ 
pounds  ? 

ANALYSIS.— The  minuend  12^=12^.  Now  i2|— sJ=6a,  Ans. 
Or,  i2i=^,  or44a;  and  5$=^.  Now^— a4a=Y.  which  reduced 
to  a  mixed  number,  equals  6J,  the  same  as  before. 

(10.)  (II.)  (12.)  (I3.)  (I4.) 

From     I  5%  7i  23f 

Take      f  3*  5f  '5*  H 

(15.)         (16.)        (17.)         (18.)         (19-) 
From    sff          yH         6f!  9$f 

Take     3#          4H          3*1          5H          7TVs 

20.  From  a  box  containing  56^  pounds  of  sugar,  a  grocer 
took  out  23!  pounds :  how  many  pounds  were  left  in  the  box? 

21.  From  a  farm  containing  165^  acres,  the  owner  sold 
78^  acres :  how  much  land  had  he  left  ? 

22.  From  13  subtract  f. 

ANALYSIS. — ist.  Reducing  13  to  sevenths,  we  have  I3=V-,  and 
s7L— 5  =  *7&,  or  12?  Ans. 

Or,  2d.  Borrowing  i  from  13,  and  reducing  it  to  ?ths,  we  have 
13=12^,  and  12}— $=12$,  Ans. 


8UBTBACTIOH  OF     FRACTIONS.             119 

23.  From  46  take  7^.  24.  From  58  take  2of. 

25.  From  84}  take  41.  2-6.  From  150^  take  83. 

27.  From  no  take  7^^-.  28.  From  1000  take  999!$ 


29.  Subtract  £  of  £  of  3  from  |  of  f  of  i^. 

ANALYSIS.—  Reducing  i       4  __i       4.    *3_3_12 

the  compound  fractions  ~  °r  7  °    I*~~2'52~~5~~2O 

to  simple  ones,  then  to  j        j.  i         I        $        i         S 

a  common  denominator,  -of  -of  3    ==  —  of-of-  =  —  =  — 

the  subtrahend  becomes  34  $414       20 

•fs,    the    minuend    ft.  ^        j^ 

—  £>  =  ?(>.    Ana.  '*  20* 

(3°-)  (3I-)  (32.)  (33-) 

From    f  of  |          fofft          A  of  4*         -^  of  28 
Take     iof-J          f  of  A  f  of3  I  of4| 


34-  From  £§  of  62-^  subtract  -^  of  i6£. 

35.  A's  farm  contains  256!  acres,  B's  431-^  acres:  what 
is  the  difference  in  the  size  of  their  farms  ? 

36.  If  from  a  vessel  containing  230-^  tons  of  coal,  119^ 
tons  are  taken,  how  much  will  there  be  left  ? 

37.  A  man  bought  a  cask  of  syrup  containing  58$ 
gallons ;  on  reaching  home  he  found  it  contained  only  1 7$ 
gallons :  how  many  gallons  had  leaked  out  ? 

38.  A  lady  having  $100,  paid  $8J  for  a  pocket  hand- 
kerchief, $15^  for  a  dress  hat,  $46!  for  a  cloak:  how 
much  had  she  left  ? 

39.  From  ™  subtract  — . 

6  5 

ANALYSIS. — Reducing  the  complex  5$_TT      ,_  TT_SS 

fractions  to  simple  ones,  then  to  a  ~g~ ^T~r()~i2      off 

common  denominator,  the  minuend  , 

becomes  f$,  the  subtrahend  £5-  (Arts,  -l^*-:-^—  -£g=   £$ 

152,    I53-)      Now,  £$  —  it  =  S3,  or  5                             "TTH is 

B,  Ant.  Ans'  etr— 4* 

40.  From  —  -  take  -A  41.  From  —  take  — . 

58  3  22 


120        MULTIPLICATION     OF     FRACTIONS. 

MULTIPLICATION    OF    FRACTIONS. 
CASE    I. 

163.  To  multiply  a  Fraction  by  a   WTiole  Number. 

Ex.  i.  What  will  4  pounds  of  tea  cost,  at  |  of  a  dollar  8 
pound  ? 

ist  METHOD. — Since  i  pound  costs  $%,  4  pounds        ist  OPERATION. 
will  cost  4  times  as  much,  and  4  times  $£  are          &ff  X4— $^* 
\a=$2^,  which  is  the  answer    required.      For,     and  $-^=$2^ 
multiplying  the  numerator,  multiplies  the  frac- 
tion.    (Art.  144  Prin.  5.) 

2d  METHOD. — If  we  divide  the  given  de-  *d  OPERATION. 

nominator  by  the  given  number  of  pounds,     $^-7-4  =  $^,  or  $2^ 
we  have  f-5-4=$f,  or  $2^,  which  is  the  true 

answer.      For,   dividing  the  denominator,   multiplies  the  fraction. 
(Art.  144,  Prin.  8.)    Hence,  the 

KULE. — Multiply  the  numerator  ~by  the  whole  number. 
Or,  divide  the  denominator  by  it. 

REMAKES. — i.  When  the  multiplicand  is  a  mixed  number,  th« 
fractional  and  integral  parts  should  be  multiplied  separately,  and 
the  results  be  united. 

Or,  the  mixed  number  may  be  reduced  to  an  improper  fraction, 
and  then  be  multiplied  as  above. 

2.  If  a  fraction  is  multiplied  by  its  denominator,  the  product  will 
be  equal  to  its  numerator.     For,  the  numerator  is  both  multiplied 
and  divided  by  the  same  number.    (Art.  86,  a.)    Hence, 

3.  A  fraction  is  multiplied  by  a  number  equal  to  its  denominator 
by  cancelling  the  denominator.     Thus,  f  x  8=5.     (Art.  121.) 

4.  In  like  manner,  a  fraction  is  multiplied  by  any  factor  of  its  de- 
nominator by  cancelling  that  factor. 

2.  Multiply  2  7 1-  by  6. 

ANALYSIS. — Multiplying  the  fraction  and  integer        OPERATION. 
Separately,  we  have  6  times  £= V»  or  3$ ;   and  6 
times  27  =  162.     Now  162  +  3^=165!,  the  answer. 
Or,  thus :  27|=H a ;  and-4*  *  6=8|»,  or  165!,  Ans.       An*.   1 65$ 

163.  How  multiply  a  fraction  by  a  whole  number  ?  Upon  what  doea  the  first 
method  depend  ?  The  second  ?  Bern.  How  multiply  a  mixed  number  by  a 
whole  one  ?  How  multiply  a  fraction  by  a  number  equal  to  its  denominator  1 
How  by  any  factor  in  ite  denominator  ? 


MULTIPLICATION     OF  FRACTIONS.         121 

(3-)  (4-)  (5-)  (6.)  (7-) 

Mult.      |f  -&  23f  35|  48f- 

By  7  9  8  10  12 


(8.)  (9.)  (10.)          (ii.)  (12.) 

fi        fi       M       9*A       85* 

By  48  100  78  26  48. 

13.  Multiply  fjf-  by  239. 

14.  What  cost  8  barrels  of  cider,  at  $7$  a  barrel  ? 

15.  At  $25!  each,  what  will  9  chests  of  tea  cost? 

16.  What  will  25  cows  come  to,  at  $48f  apiece? 

17.  What  cost  27  tons  of  hay,  at  $29^  a  ton  ? 

1  8.  What  cost  35  acres  of  land,  at  $45  £  per  acre  ? 

19.  At  $34!:  apiece,  what  cost  50  hogsheads  of  sugar. 

20.  At  $45  1  apiece,  what  will  100  coats  come  to  ? 

21.  What  cost  6  dozen  muffs,  at  $75^  apiece  ? 

CASE  II. 
165.  To  Multiply  a   Wliole  Number  by  a  Fraction. 

DBF.  Multiplying  by  a  Fraction  is  taking  a 
certain  part  of  the  multiplicand  as  many  times  as  there 
are  like  parts  of  a  unit  in  the  multiplier. 

But  to  find  a  given  part  of  a  number,  we  divide  it  into 
as  many  equal  parts  as  there  are  units  in  the  denominator, 
and  then  take  as  many  of  these  parts  as  are  indicated  by 
the  numerator.  (Art.  143.)  That  is, 

To  multiply  a  number  by  £,  divide  it  by  2. 

To  multiply  a  number  by  %,  divide  it  by  3. 

To  multiply  a  number  by  f,  divide  it  by  4  for  £,  and 
multiply  this  quotient  by  3  for  |,  etc.  Hence, 

REMARKS.  —  r.  Multiplying  a  whole  number  by  a  fraction  i? 
the  same  as  finding  a  fractional  part  of  a  number,  which  the  pupil 
dhould  here  review  with  care.  (Art.  143.) 

Dtf.  What  is  it  to  multiply  by  a  fraction  ?   How  multiply  by  1  •   By  }  ?  By  J  f 


122  MULTIPLICATION    OF    FRACTIONS. 

2.  When  the  multiplier  is  i,  the  product  is  equal  to  the  multipli- 
cand. 

When  the  multiplier  is  greater  than  i,  the  product  is  greater  than 
t.ie  multiplicand. 

When  the  multiplier  is  less  than  i,  the  product  is  less  than  the 
multiplicand. 

1.  What  will  £  of  a  ton  of  iron  cost,  at  855  a  ton  ? 

ist  METHOD. — Sinco  i  ton  costs  §55,  |  of  a  ton  OPERATIC*. 

will  cost  £  times  $55,  or  £  of  $55.    But  J  of  $55  $4)55 

=  3  times  £  of  $55.  Now  \  of  $55=855-^-4,  or 
$13^  ;  and  3  times  $I3$=$4ll,  the  answer  required. 
For,  by  definition,  multiplying  by  a  fraction  is  Ane 
taking  apart  of  the  multiplicand  as  many  times  as 
there  are  like  parts  of  a  unit  in  the  multiplier.  Now  dividing  the 
Avhole  number  by  4,  takes  i  fourth  of  the  multiplicand  once ;  and 
multiplying  this  result  by  3,  repeats  this  part  3  tunes,  as  indicated  by 
the  multiplier.  (Art.  143.) 

2d  METHOD. — }  of  $55=i  of  3  times  $55.    But  $rt 

$55  x  3  =  $165,   and   $i6s-:-4— $41^,  the   same  as 
before.     For,  i  of  3  times  a  number  is  the  eame  as  \    /•    " 

3  times  i  of  it.     Hence,  the 

Ans.  $414 

RULE. — Divide  the  whole  number  by  the  denominator  of 
the  fraction,  and  multiply  by  the  numerator. 

Or,  Multiply  the  ivhole  number  by  the  numerator  of  the 
fraction,  and  divide  by  the  denominator. 

REMARKS. — i.  The  fraction  may  be  taken  for  the  multiplicand, 
and  the  whole  number  for  the  multiplier,  at  pleasure,  without  affect- 
ing  the  result.  (Art.  45.) 

2.  When  the  multiplier  is  a  mixed  number,  multiply  by  the  frac- 
tional and  integral  parts  separately,  and  unite  the  results. 

Or,  reduce  it  to  an  improper  fraction ;  then  proceed  according  to 
the  rule.  (Ex.  2.) 

2.  What  will  8|  tons  of  copper  ore  cost,  at  $365  a  ton  ? 
ANALYSTS. — 8}  =*$•  ;  now  46a  tons  will  come  to  *<*  times  $365  ;  and 

$365  *"  =  $3 1 39-  An*. 


165.  How  is  a  whole  number  multiplied  by  a  fraction  ?  Hem.  What  is  the 
operation  like  ?  When  the  multiplier  is  i,  what  is  the  prodnct  ?  W  hen  the  mul- 
tiplier is  greater  than  i,  what  f  When  less,  what?  Hem  How  multiply  a  whola 
by  a  mixed  number  ? 


MULTIPLICATION     OF     FRACTIONS.         123 

(30          (4.)           (5-)  (6.)  (7.) 

Mult.        65  89  96  87  100 

By         j  &  _J|         _4j  _5* 

8.  What  cost  T7ff  of  an  acre  of  land,  at  $38  an  acre  ? 

9.  At  $29  a  ton,  what  cost  8|  tons  of  hay  ? 

10.  What  cost  28£  tons  of  lead,  at  $223  per  ton? 

n.  What  is  fj'of   845?  15.  Multiply   79  by  7$. 

12.  What  isT7^  of  1876  ?  16.  Multiply  103  by  9$. 

13.  What  is   |4  of  1000?  17.  Multiply  1001  by  21-^. 

14.  What  is  £££  of  2010  ?  18.  Multiply  1864  by  37^. 

CASE    III. 
166.  To  multiply  a  Fraction  by  a  Fraction. 

i.  What  will  |  of  a  pound  of  tea  cost,  at  $^  a  pound  ? 

ANALYSIS. — £  of  a  pound  will  cost  I  sixth  OPERATION. 

as  much  as  i  pound ;  and  £  of  $,90-  is  $6V  "&  x  f~¥§»  or  *l 
Again,  §  of  a  pound  will  cost  5   times  as          BY  CANCELLATION. 

much  as  £  ;  and  5  times  $&  are  $$8=$J,  3>  j^_  x  5,  i_  . . 

the  answer  required.  2,  IE       6,  2 

Or,  having  indicated  the  multiplication,  caned  the  common  fac- 
tors 3  and  5,  and  then  multiply  the  numerators  together  and  the  de- 
nominators ;  the  result  is  $J,  the  same  as  before 

The  reason  is  this:  Multiplying  the  denominator  of  the  multipli- 
cand ,90  by  6  the  denominator  of  the  multiplier,  the  result,  $,&,,  is 
5  times  too  small ;  for,  we  have  multiplied  by  £  instead  of  £  of  a 
unit,  the  true  multiplier.  To  correct  this,  we  must  multiply  this  re- 
sult by  5,  which  is  done  by  multiplying  its  numerator  by  5.  (Art.  144.) 

The  object  of  cancelling  common  factors  is  twofold:  it  shortens  the 

operation,  and  gives  the  answer  in  the  lowest  terms.    Hence,  the 

/ 
KULE. — Cancel  the  common  factors  ;  then  multiply  the 

numerators  together  for  the  new  numerator,  and  the  denom- 
inators/or the  new  denominator.     (Art.  144,  Prin-  9.) 

166.  How  multiply  a  fraction  by  a  fraction  ?  Object  of  cancelling  ?  Ecm.  How 
multiply  compound  fractions  ?  Mixed  numbers !  Complex  fractions  ? 


124       MULTIPLICATION     OF     FRACTIONS. 

REMARKS. — i.  Mixed  numbers  should  be  reduced  to  improper 
fractions,  and  complex  fractions  to  simple  ones ;  then  apply  the  rule. 

"2.  Multiplying  by  a  fraction,  reducing  a  compound  fraction  to  a 
timpte  one,  and  finding  a  fractional  part  of  a  number,  are  identical 
operations.  (Arts.  143,  149,  166.) 

2.  Multiply  f  of  |  of  £  by  \  of  f. 

SOLUTION. — It  is  immaterial  as  to  the  result,  whether  $  2 

the  fractions  are  arranged  in  a  horizontal  or  in  a  per-  24, 

pendicular  line,  with  the  numerators  on  the  right  and  6  5 

the  denominators  on  the  left.    (Art.  122,  Rent.)  3.  i 

J 

Perform  the  following  multiplications : 

3-      T  X     TJT  7-    T¥  X  TJ  II.      if  X     -y  X     ^ 

41    TTf  X  T5"  "•   TT  X  yy  12.      -g-  X  -j-jj-  X     ^- 

15.  Multiply  f  of  ^  of  25  by  f  of  ^  of  f . 

16.  Multiply  |  of  A  of  33i  by  £  of  ^  of  i8|. 

17.  If  i  quart  of  cherries  costs  \  of  f  of  45  cents,  what 
will  |  of  A  of  a  quart  cost  ? 

1 8.  What  will  5^  barrels  of  chestnuts  cost,  at  6f  dollars 
a  barrel  ?  Ans.  $36. 

(19.)  (20.)  (21.)  (22.) 

Mult.          24^  37^  62^  165^ 

By  84  i8£  31*  92! 


23 


el  2\ 

What  is  product  of  ~  multiplied  by  — . 


21      .     21         2|    8          8 
-  --=.-•  -7-3  =  -. 


. 
ANALYSIS.  —  ,-=  —  i-6  =  —  .and 

64  24*         3 

7,21     8.  7        7 

Cancelling  common  factors,  \re  have        —  X  -      =  ---  =  -.Ana. 

3>2*     9,3     3x3     9' 

24.  Multiply  ?i  by  ^.  25.  Multiply  ^  by  ^. 


MULTIPLICATION     OF     FRACTIONS.        125 

167.  The  preceding  principles  may  be  reduced  to  the 
following 

GENERAL    RULE. 

I.  Reduce  ivhole  and  mixed  numbers  to  improper  frac- 
tions, and  complex  fractions  to  simple  ones. 

II.  Cancel  the  common  factors,  and  place  the  product  of 
the  numerators  over  the  product  of  the  denominators. 

EXAMPLES. 

1.  What  cost  £  of  a  yard  of  calico,  at  $£  a  yard  ? 

2.  "What  cost  f  of  a  pound  of  nutmegs,  at  $|  a  pound  ? 

3.  At  $f§  a  gallon,  what  will  £  of  a  gallon  of  molasses 
cost? 

4.  Multiply  23!  by  6\.         7.  Multiply  pi-f  by  43$. 

5.  Multiply  42T2T  by  8f       8.  Multiply  i64|  by  75^. 

6.  Multiply  65!  by  gf.          9.  Multiply  2oof  by  86T7^. 

10.  What  is  the  product  of  f  of  £  of  £  of  f  into  y  of  -^ 


1  1.  What  is  the  product  of  \  of  3^  into  -|  of  19^  ? 

12.  When  freight  is  $i|  per  hundred,  what  will  it  cos4 
to  transport  27  hundredweight  of  goods  from  New  York 
to  Chicago  ? 

13.  What  will  1  6£  quarts  of  strawberries  come  to,  at 
26^  cents  per  quart  ? 

14.  At  $3^  a  barrel,  what  will  be  the  cost  of  47!  barrels 
of  cider? 

15.  Bells  are  composed  of  ^  tin  and  f  copper:  how 
many  pounds  o.  each  metal  does  a  church-bell  contain 
which  weighs  3750  pounds  ? 

1  6.  What  will  45  men  earn  in  15^  days,  if  each  earns 
$2f  per  day? 

17.  How  many  feet  of  boards  will  a  fence  603-^  rods 
long  require,  allowing  74^  feet  of  boards  to  a  rod  ? 

167.  What  Is  the  general  rule  for  multiplying  fractions  ? 


126  DIVISION     OF     FRACTIONS. 

DIVISION    OF    FRACTIONS. 

CASE   I. 
168.  To  divide  a  Fraction  by  a  Wliole  Number. 

1.  If  4  yards  of  flannel  cost  $f,  what  will  i  yard  cost? 
ist  METHOD.  —  i  yard  will  cost  i  fourth  as 

much  as  4  yards  ;  and  dividing  the  numerator 

by  4,  we  have  $§-*-4=$f,  the  answer  required.      ^---4.  __  ^2  ^^ 

For,  dividing  the  numerator  by  any  number,         9 

divides  the  fraction  by  that  number. 

2d  METHOD.  —  Multiplying  the  denominator  ad  METHOD. 

by  the   divisor  4  (the  number  of  yards),  the 
result  is  -3^  =  f  ,  the  same  as  before.     For,  0x4       ^6 

multiplying  the  denominator  by  any  number,  g 

divides  the  fraction  by  that  number.    (Art.  144,          Ans.  —  =•  $f 
Prin.  7.)  .  Hence,  the 

EULE.  —  Divide  the  numerator  by  the  whole  number. 
Or,  multiply  the  denominator  by  it. 

REMARK.  —  When  the  dividend  is  a  mixed  number,  it  should  be 
reduced  to  an  improper  fraction  ;  then  apply  the  rule. 

2.  Divide  $f  by  9.     Ans.  $f  -^9=$^. 

Perform  the  following  divisions  : 
3. 
4-  U-9-  8.  W-f-67.  12. 


5. 

6. 


9. 


14. 


15.  A  man  having  |  of  a  barrel  of  flour,  divided  it 
equally  among  5  persons  :  what  part  of  a  barrel  did  each 
receive  ? 

16.  If  12  oranges  cost  $TVb>  what  will  i  orange  cost? 

17.  If  7  writing-books  cost  $|,  what  will  that  be  apiece? 

18.  If  6  barrels  of  flour  cost  $45  f,  what  will  i  barrel  cost  ? 


168.  How  divide  a  fraction  by  a  whole  number  ?  Upon  what  does  the  first 
method  depend?  The  second?  Which  is  preferable?  Bern.  How  is  a  mixed 
number  divided  by  a  whole  one  ? 


DIVISION     OF     FRACTIONS. 


127 


Perform  the  following  divisions : 

19.  8oT4j-^i2.  23.  865tV-S2.          27. 

20.  763^-^  J4-  24.  490/^—40.  28.  46841^68. 

21.  284--f-2O.  25.    758^f —  48.  29.    7896-1-7-25. 

22.  5lf  4-27.  26.    975iiy-7-63.  30.    9684^-^-84. 

31.  If, 5  yards  of  cloth  cost  $42 f,  what  will  i  yard  cost. 

32.  If  6  horses  cost  $756^,  what  will  i  horse  cost? 

33.  If  45  Ibs.  of  wool  cost  $54|,  what  will  i  Ib.  cost  ? 


CASE   II. 
168.  «•    To  divide  a  Whole  If  umber  by  a  Fraction. 

i.  At  $|  a  yard,  how  many  yards  of  cashmere  can  you 
buy  for  $20  ? 


ANALYSIS. — At  $V  a  yard,  $20  will  OFEBATTOIT. 

buy  as    many  yards    as    there    are  20x4=80  y.  at  $£. 

fourths  in  $20,  and  4  times  20  are  80.  80-1-3  =  26^  y.  at  $J. 

But  the  price  $J,  is  3  times  aa  much  Or,  20  X  $=26$  y.  at  $J. 
as  $\;  therefore,  at  $J,  you  can  buy 

only  i  as  many  yards  as  at  $| ;  and  ^  of  80  yards  is  26 }  yards, 
the  answer  required.  For,  multiplying  the  whole  number  by 
the  denominator  4,  reduces  it  to  fourths,  which  is  the  same  denomi- 
nator as  the  given  divisor.  But  when  fractions  have  a  common 
denominator,  their  numerators  are  like  numbers;  therefore,  onw 
may  be  divided  by  the  other,  as  whole  numbers.  (Art.  156.)  But 
multiplying  the  whole  number  20,  by  the  denominator  4,  and  dividing 
the  product  by  the  numerator  3,  is  the  same  as  inverting  the 
fractional  divisor,  and  then  multiplying  the  dividend  by  it.  Hence,  the 

RULE. — Multiply  the  whole  number  by  the  fraction  in- 
verted.    (Art  165.) 

REMARKS. — i.  When  the  divisor  is  a  mixed  number  it  should  be 
reduced  to  an  improper  fraction  ;  then  divide  by  the  rule.    (Ex.  16.) 

2  A  fraction  is  inverted,  when  its  terms  are  made  to  exchange 
places.    Thus,  $  inverted  becomes  $. 

3  After  the  denominator  is  inverted,  the  common  factors  shoulJ 
be  cancelled.    (Art.  149,  n.) 


168,  a.  How  divide  a  whole  number  by  a  fraction  T    How  does  it  appear  that 
this  process  will  give  the  true  answer? 


128  DIVISION     OF     FBACTION8. 

Perform  the  following  divisions : 

2-  95-^1-  5-  175-^-nr-  8.  576-1-^. 

3.  168-=-^.  6.  261-*-^.  9.  1236-^- 

4.  245-^.  7.  348-7-3..  10.  6240-^-^. 

ii.  How  many  yards  of  muslin,  at  $|  a  yard,  can  be 
bought  for  $19  ? 

ANALYSIS. — At  $^  a  yard,  $19  will  buy  as  many  yards  as  there 
are  thirds  in  19,  and  19  x  $=57.  Therefore  $19  will  buy  57  yards. 

12.  27  =how  many  times  £  ?     14.  38=how  many  times  |? 

13.  53 = how  many  times  £?     15.  67  =how  many  times  -j^? 

1 6.  How  many  cloaks  will  45  yards  of  cloth  make,  each 
containing  4^  yards  ? 

ANALYSIS. — Reducing  the  divisor  to  an  improper  fraction,  we  have 
$}—%,  and  45  yds.  x  f=99a,  or  10 cloaks.,  An». 

17.  Divide  88  by  lof.  19.  Divide  785  by  62-}. 

18.  Divide  100  by  12^.  20.  Divide  1000  by  87^. 

CASE   III. 

169.  To  divide  a  Fraction  by  a  Fraction,  when  they  have  a 
Common,  Denominator. 

REMARK. — This  case  embraces  two  classes  of  examples: 
First,  those  in  which  the  fractions  have  a  common  denominator. 
Second,  those  in  which  they  have  different  denominators. 

2 1.  At  $f  apiece,  how  many  melons  can  be  bought  for  &£  ? 

ANALYSIS. — Since  $|  will  buy  i  melon,  $£  will  buy    OPERATION. 
as  many  as  $|  are  contained  times  in  $J  ;    and  3     ^~J~Y:=::1 
eighths  are  in  7  eighths,  2  times  and  i  over,  or  2^  times.  Ans.  2$  m. 
(Art.  156.)     Hence,  the 

RULE. — Divide  the  numerator  of  the  dividend  by  that 
of  the  divisor. 

NOTE.— When  two  fractions  have  a  common  denominator,  their 
numerators  are  like  numbers,  and  the  quotient  is  the  same  as  if  they 
were  whole  numbers.  (Arts.  64,  156.) 

22.  Divide  ff  by  ~fa.  24.  Divide  f|  by  £J. 

23.  Divide  ff  by  jj. 25.  Divide  j|  by  |f 

169.  How  divide  a  fraction  by  a  fraction  when  they  have  a  common  denominator  t 


DIVISION    OF    FRACTIONS.  12S 


170.  To  divide  a  Fraction  by  a  Fraction,  when  they  have 
Different  Denominators. 

1.  At  $|  a  pound,  how  much  tea  can  be  bought  for  $£  ? 
IST  ANALYSIS.  —  $f  will  buy  as  many  pounds  ^=.^=~^ 

as  $J  are  contained  times  iu  $f.     $£=$*.     Re-  — 

ducing  J  and  $  to  a  common  denominator,  they 
become  -&  and  -,%,and  their  numerators  like  num- 
fors.  Hence,  -&-5--&=9-s-  8,  or  i\.  Am.  ii  pound. 

REMARKS.  —  i.  In  reducing  two  fractions  to  a  common  denomina- 
tor, we  multiply  the  numerator  of  each  into  the  denondnator  of  the. 
other,  and  the  two  denominators  together.  But  in  dividing,  no  use 
ia  made  of  the  common  denominator  ;  hence,  in  practice,  multiply- 
ing the  denominators  together  may  be  omitted.  (Art.  153.) 

20  ANALYSIS.—  $§=$!•  At  $^  a  pound,  $i  will  buy  as  many 
pounds  as  $}  are  contained  times  in  $i.  Now  i  -*-*=$-*-*,  and 
?,-*-$=  |,  or  f  pounds,  the  quotient  being  the  divisor  inverted.  Again, 
if  $i  will  buy  J  pounds,  $f  will  buy  $  of  f  pounds  ;  and  I  of  i  =  £, 
Or  i  J  pounds,  the  same  as  before. 

REMARKS.  —  2.  In  this  analysis,  it  will  be  seen,  by  inspection,  that 
the  numerator  of  each  fraction  is  also  multiplied  into  the  denomi- 
nator of  the  other.  Both  of  these  solutions,  therefore,  bring  the 
same  combinations  of  terms,  and  the  same  result,  as  inverting  the  di- 
visor, and  multiplying  the  dividend  by  it.  (Art.  166.)  Hence,  the 

KULE.  —  Reduce  the  fractions  to  a  com.  denominator,  and 
divide  the  numerator  of  the  dividend  by  that  of  the  divisor. 
Or,  multiply  the  dividend  by  the  divisor  inverted. 

NOTES.  —  i.  The  first  method  is  based  upon  the  principle  that 
numerators  of  fractions  having  a  com.  denom.  are  like  numbers. 

The  second  is  evident  from  the  fact  that  it  brings  the  same  com- 
fcinations  as  reducing  the  fractions  to  a  common  denominator. 

The  divisor  is  inverted  for  convenience  in  multiplying, 

2.  After  the  divisor  is  inverted,  the  common  factors  should  be 
cancelled,  before  the  multiplication  is  performed. 

3.  Mixed  numbers  should  be  reduced  to  improper  fractions,  com- 
pound and  complex  fractions  to  simple  ones. 

4.  Those  expressions  which  have  fractional  denominators,  are  re- 
reduced  to  simple  fractions  by  the  above  rule.  Ex.  14.  (Art  141.  Rem.) 

170.  How  when  they  have  not?  Note.  Upon  what  principle  is  the  first  method 
based  ?  The  second  ?  Show  this  coincidence.  What  ie  done  with  mixed  num- 
bers, compound,  and  complex  fractions. 


130 


DIVISION     OF     FRACTIONS. 


Perform  the  following  divisions  : 


2. 


jr 3 


4.    -fa—-*?. 

5-  r¥~~A* 


6. 

7- 
8- 
9- 


27! 


14.  Reduce  — ^  to  a  simple  fraction. 
i  of 


SOLUTION. — The  given  expression  is  equivalent  to  275-^-10^,  and  ia 
reduced  to  a  simple  fraction  by  performing  the  division  indicated. 
Ans.  f  or  2\.  (Art.  141,  Rem.) 

is4 

1 7.  Reduce  -~^ 

I  I-J  1 2O-5- 

16.  Reduce  — T-  18.  Reduce  - 


15.  Reduce  - 


19.  Divide  777^  dollars  by  129!  dollars. 

20.  How  many  rods  in  2320^  feet,  at  16^  feet  to  a  rod  ? 

21.  How  many  times  30^  sq.  yards  in  320^  sq.  yards? 

22.  What  is  the  quotient  of  f  off  of  4}  divided  by  f  of  f  ? 

23  25 

SOLUTION. — -  of  -  of  44  divided  by  -  of  „  = 
34  7s 


2      2      5         5 

When  the  perpendicular  form  is  adopted,  the 
divisor  must  bj  inverted,  before  its  terms  are  arranged. 

23.  Divide  f  of  f  of  £  by  £  of  f 

24.  Divide  £  of  f  of  |  by  £  of  f  . 

25.  Divide  f  of  f  of  4^  by  £  of  |  of  4$. 

i  g 

26.  What  is  the  quotient  of  f  divided  by  -=• 

f  3i 

J 

ANALTSIS.—  The  dividend  |  = 


Q 

x  |  ;  the  divisor  -v 


-v 


Q 

=  -  X 


16 


XT  O  T 

27.  What  is  quotient  of  ~  divided  by  --  ? 
^r  rr 


DIVISION     OF     FRACTIONS.  131 

171.  The  preceding  principles  may  be  reduced  to  the 
following 

GENERAL    RULE. 

Reduce  ivhole  and  mixed  numbers  to  improper  fractions, 
compound  and  complex  fractions  to  simple  ones,  and  mul- 
tiply the  dividend  by  the  divisor  inverted. 

NOTE.  —  After  the  divisor  is  inverted,  the  common  factors  should 
be  cancelled. 

EXAMPLES. 

1.  If  a  young  man  spends  $2^  a  month  for  tobacco,  in 
what  time  will  he  spend  $13^  ?     (Art.  48,  Note  3.) 

2.  If  a  family  use  5^  pounds  of  butter  a  week,  how  long 
will  45  1  pounds  last  them  ? 

3.  If  $|;  will  buy  i  yard  of  gingham,  how  much  will 


4.  How  many  tons  of  coal,  at  $7^  a  ton,  can  be  bought 
for  $125!? 

5.  How  many  times  will  a  keg  containing  13!  gallons 
of  molasses  fill  a  measure  that  holds  f  of  a  gallon  ? 

6.  What  is  the  quotient  of  24-^  divided  by  8f  ? 

7.  What  is  the  quotient  of  ff  divided  by  H  ? 

8.  What  is  the  quotient  of  f££  divided  by  $f  ? 

9.  What  is  the  quotient  of  45!  divided  by  25^? 

10.  How  much   tea,  at  $i|-  a  pound,  can  be  bought 
for$75f? 

11.  How  many  acres  can  be  sowed  with  57^  bushels  of 
oats,  allowing  i|  bushel  to  an  acre?. 

12.  A  man  having  57^  acres  of  land,  wished  to  fence  it 
into  lots  of  5^  acres  :  how  many  lots  could  he  make  ? 

13.  How  many  yards  of  cloth,  at  $6|,  can  yon  buy  for 


14.  What  is  the  quotient  of  f  of  f  of  If  -=-|  of  |  of  z\  ? 

15.  What  is  the  quotient  of  f  of  |  of  8f  -=-4f  }  ? 

171.  What  is  the  general  rule  for  dividing  fractions  t 


132  DIVISION     OF     FRACTIONS. 


QUESTIONS     FOR     REVIEW. 

1.  A  book-keeper  adding  a  column  of  figures,  made  the 
I3sulfc  $563!;  proving  his  work,  he  found  the  true  amount 
to  be  $607^ :  how  much  was  the  error  ? 

2.  A  merchant  paid  two  bills,  one   $278!,  the  other 
$34o|,  calling  the  amount  $638^ :  what  should  he  have 
paid  ?     What  the  error  ? 

3.  What  is  the  sum  of  35^  and  23!  minus  8$? 

4.  A  speculator  bought  two  lots  of  land,  one  containing 
47-f3^  acres,  the  other  63*  acres:  after  selling  78$  acres, 
how  many  had  he  left  ? 

1 2  5 

5.  What  is  the  sum  of  -^  plus  •—  ? 

3*          3t 

6.  What  is  the  sum  —  plus  —  plus  — r  ? 

4  *        if  *       5i 

7.  A  man  owning  f£  of  a  ship  worth  $48064,  sold  £  of 
his  share.     What  part  of  the  ship  did  he  sell ;  what  part 
does  he  still  own,  and  what  is  it  worth  ? 

8.  A  farmer  owning  75!  acres,  sold  31^  acres,  and  after- 
ward bought  42!  acres :  how  many  acres  did  he  then  have  ? 

3  .5 

o.  What  is  the  difference  between  ~  and  ~  ? 

I          6 
6  s  ?- 

10.  What  is  the  difference  between  — =-  and  *?.  ? 

3i          o 

11.  What  is  the  difference  between  -~  and  -  ;v  ? 

12.  If  it  requires  i-J  bushels  of  wheat  to  sow  an  acre, 
how  many  bushels  will  be  required  to  sow  28$  acres  ? 

13.  How  many  feet  in  148$  rods,  allowing  16$  feet  to 
a  rod? 

14.  If  a  pedestrian  can  walk  45^  miles  in  i  day,  how 
far  can  lie  walk  in  i8f  days  ? 

15.  What  will  37$  barrels  of  apples  come  to,  at  $2^  per 
barrel ? 


QUESTIONS     FOB     EEVIEW.  133 

1  6.  The  sum  of  two  numbers  is  68||,  and  the  difference 
between  them  is  13!  :  what  are  the  numbers? 

17.  What  is  the  product  of  -f  into  -f  ? 

3*          4i 

1  8.  What  is  the  product  of  ^  into  ^? 

6|          12 

19.  What  is  the  product  of  —I  into  ?f  ? 

I2|  2\ 

20.  How  many  days'  work  will  100  men  perform  in  \$ 
of  a  day  ? 

21.  A  man  owning  -^  of  a  section  of  land,  sold  -J  of  his 
share  for  $izf  :  what  is  the  whole  section  worth,  at  that 
rate? 

22.  How  many  times  is  -^  of  f  of  5  J  contained  in  23!  ? 

23.  Divide  £  of  i8|  by  f  of  f  £  of  f  of  31$. 

24.  If  a  gang  of  hands  can  do  ^f  of  a  job  in  5!  days, 
what  part  of  it  can  they  do  in  i  day  ? 

25.  If  |  of  a  yard  of  satin  will  make  i  vest,  how  many 
vests  can  be  made  from  31^  yards  ? 

26.  How  many  oil-cans,  each  containing  if  gallon,  can 
be  filled  from  a  tank  of  6  if  gallons  ? 

27.  If  a  man  walks  3^  miles  an  hour,  how  long  will  it 
take  him  to  walk  45^  miles  ? 

28.  By  what  must  Jf  be  multiplied  to  produce  15!? 

29.  How  many  bushels  of  apples,  at  f  of  a  dol.,  are 
required  to  pay  for  6  pair  of  boots,  at  $6  J  ? 

30.  A  farmer  sold  330^  pounds  of  maple  sugar,  at  i6| 
cents  a  pound,  and  took  his  pay  in  muslin,  at  22^  cents  a 
yard:  how  many  yards  did  he  receive? 

31.  Divide  the  quotient  of  12^  divided  by  3^  by  the 
quotient  of  6£-=- 


32.  What  is  the  quotient  of  -f  -r  §? 

¥          TT 

•5  I. 

33.  What  is  the  quotient  of  12^  times  =•* 

4 


134  FRACTIONAL     RELATION 


FRACTIONAL    RELATION    OF    NUMBERS. 

172.  That  Numbers  may  bo  compared  with  each  other  frac- 
tionally, they  must  be  so  far  of  the  same  nature  that  one  may  prop 
erly  be  said  to  be  a  part  of  the  other.     Thus,  an  inch  may  be  com- 
pared with  afoot ;  for  one  is  a  twelfth  part  of  the  other.     But  it  can- 
not be  said  that  &foot  is  any  part  of  an  hour  ;  therefore  the  former 
cannot  be  compared  with  the  latter. 

173.  To  find  what  part  one  number  is  of  another. 

1.  What  part  of  4  is  i  ? 

ANALYSIS. — If  4  is  divided  into  4  equal  parts,  one  of  those  parts  is 
called  i  fourth.     Therefore,  i  is  \  part  of  4. 

2.  What  part  of  6  is  4  ? 

ANALYSIS. — i  is  £  of  6,  and  4  is  4  times  £,  or  $  of  6.    But  f =$ 
(Art.  146) ;  therefore,  4  is  f  of  6.     Hence,  the 

RULE. — Make  the  number  denoting  the  part  the  numera- 
tor, and  that  with  which  it  is  compared  the  denominator. 

NOTE. —  i.  This  rule  emb races  four  classes  of  questions : 
i  st.  What  part  one  whole  number  is  of  another. 
2d.  What  part  &  fraction  is  of  a  whole  number. 
3d.  What  part  a  whole  number  is  of  a  fraction. 
4th.  What  part  one  fraction  is  of  another. 

2.  When  complex  fractions  occur,  they  should    be  reduced  to 
timple  ones,  and  all  answers  to  the  lowest  terms.    (Art.  146.) 

3.  What  part  of  75  is  15  ?     Of  84  is  30  ? 

4.  Of  91  is  63?      6.  Of   8iisi8?      8.  Of  256  is    72? 

5.  Of  48  is  72  ?       7.  Of  ioo  is  75  ?      9.  Of  375  is  425  ? 

10.  What  part  of  i  week  is  5  days  ? 

11.  A  man  gave  a  bushel  of  chestnuts  to  17  boys:  what 
part  did  5  boys  receive  ? 

12.  At  $13  a  ton,  how  much  coal  can  be  bought  for  $10  ? 

13.  A  father  is  51  years  old,  and  his  son's  age  is   17: 
what  part  of  the  father's  age  is  the  son's  ? 


OF     NUMBERS.  135 

14.  If  8  pears  cost  35  cents,  what  will  5  pears  cost  ? 

ANALYSIS.  —  5  pears  are  $  of  8  pears  ;  hence,  if  8  pears  cost  35 
ceats,  5  pears  will  cost  3  of  35  cents.  Now  i  of  35  cents  is  4!  cents, 
and  5  eighths  are  5  times  4  2  cents,  which  are  zil  cents. 

15.  If  5  bar.  of  flour  cost  $45,  what  will  28  bar.  cost? 

16.  If  50  yds.  of  cloth  cost  $175,  what  will  17  cost? 

17.  If  25  bu.  of  apples  cost  $30,  what  will  no  bu.  cost? 
r8.  What  part  of  5  is  f  ? 

ANALYSIS.  —  Making  the  fraction  which  denotes  the  part  the  nu- 
merator, and  the  whole  number  the  denominator,  we  have  a  fraction 
to  be  divided  by  a  whole  number.  For,  all  denominators  may  be  con- 
sidered as  divisors.  Thus,  J-5-5=iau>  Ana.  (Art.  142.) 

19.  What  part  of  25  is  |?        21.  What  part  of  30  is  -^  ? 

20.  What  part  of  35  is  -1/  ?      22.  What  part  of  40  is  -^  ? 

23.  If  5  acres  of  land  cost  $100,  what  will  £  acre  cost  ? 

ANALYSIS.  —  i  acre  is  £  of  5  acres,  and  $  of  an  acre  is  J  of  £,  or  -fa 
of  5  acres.  Hence,  5  of  an  acre  will  cost  .ft,  of  $100.  Now  ^  of  $100 
is  $5  ;  and  3  twentieths  are  3  times  5  or  $15. 

24.  When  coal  is  $95  for  15  tons,  what  will  f  ton  cost? 

25.  If  19  yards  of  silk  cost  $60,  what  will  £  yard  cost? 

26.  What  part  of  f  is  2  ? 

ANALYSIS.  —  Making  the  whole  number  which  denotes  the  part,  the 
numerator,  and  the  fraction  the  denominator,  we  have  a  whole  num- 
ber to  be  divided  by  a  fraction.  Thus,  2-7-^=2  x  ^=13a,  Ans. 

27.  What  part  of  £  is  8  ?  29.  What  part  of  f  is  1  1  ? 

28.  What  part  of  |  is  12  ?          30.  What  part  of  T7V  is  20  ? 
3  1  .  What  part  of  f  is  f  ? 

ANALYSIS.  —  Making  the  fraction  denoting  the  part  the  numerator, 
and  the  other  the  denominator,  we  have  a  fraction  to  be  divided  by 
a  fraction.  Thus,  $-=-5  =  !  x  $-=$,  Ans.  (Art.  170.) 


32.  What  part  of  f  is  $  ?  34.  What  part  of  |4  is 

33.  What  part  of  ff  is  H  ?        35.  What  part  of  H  is  |-f  ? 
36.  6£  is  what  part  of  25  ? 

ANALYSIS.  —  Reducing  the  mixed  number  to  an  improper  fraction. 
we  have  &i  =  V.  and  V^-25=:i,  Ans. 


130  FRACTIONAL     RELATION. 

37.  What  part  of  100  is  12^?  40.  Of  100  is  62^? 

38.  What  part  of  100  is  33^?  41.  Of  100  is  i8J? 

39.  What  part  of  100  is  i6f  ?  42.  Of  100  is  87^? 
43.  12^  is  what  part  of  iSf  ? 

ANALYSIS.  —  Reducing  the  mixed  numbers  to  improper  fractions 
we  have,  i2^=2/,  and  iS}=1f-.    Now  2ai-j-14i=f,  Ans. 


44.  What  part  of  62^  is  i8|  ?          45.  Of  87^  i 

46.  At  $£g  a  pound,  how  much  tea  will  $f  buy  ? 

47.  At  $££  per  foot,  how  many  feet  of  land  can  be 
bought  for  *f|  ? 

48.  A  lad  spent  i8f  cents  for  candy,  which  was  62^ 
cents  a  pound  :  how  much  did  he  buy  ? 

49.  A  can  do  a  certain  job  in  8  days,  and  B  in  6  days  ' 
what  part  will  both  do  in  i  day  ? 

50.  What  part  of  4  times  20  is  9  times  16  ? 

51.  What  part  of  75  x  18  is  105-7-25  ? 

52.  What  part  of  (68  —  24)  x  14  is  168  H-  12  ? 

174.  To  find  a  Number,  a  Fractional  Part  of  it  being  given. 

Ex.  i.  9  is  £  of  what  number  ? 

ANALYSIS.  —  Since  9  is  i  third,  3  thirds  or  the  whole  number  must 
be  3  times  9  or  27.     Therefore,  9  is  a  third  of  27. 
Or,  thus  :  g  is  £  of  3  times  9,  and  3  x  9=27.    Therefore,  etc. 

2.  2  1  is  f  of  what  number  ? 

ANALYSIS.  —  Since  |  of  a  certain  number  is          OPERATION. 
21  units,  %  or  the  whole  number  must  be  as  many      21-7-^=21  X^ 
units  as  J  are  contained  times  in  21  ;  and  2i-i-J=      21x^=28. 
21  x  $=28,  the  answer  required.     For,  a  whole 
number  is  divided  by  a  fraction  by  multiplying  the  former  by  the 
latter  inverted.     (Art.  168,  a.) 

Or,  thus  :  Since  21  is  5  of  a  certain  number,  I  fourth  of  it  is  I  third 
of  21,  or  7.  Now  as  7  is  i  fourth  of  the  number,  4  fourths  must  be 
4  times  7  or  28,  the  same  as  before.  Hence,  the 

RULE.  —  Divide  the  number  denoting  the  part  by  the 
fraction. 

Or,  Find  one  part  as  indicated  by  the  numerator  of  tM 
fraction,  and  multiply  this  by  the  denominator. 


OF     KUMBEKS.  137 

NOTE.  —  The  learner  should  observe  the  difference  between  finding 
i  of  a  number,  when  J  or  the  whole  number  is  given,  and  when  only 
£  or  a  part  of  it  is  given.  In  the  former,  we  divide  by  the  denomi- 
nator of  the  fraction  ;  in  the  latter,  by  the  numerator,  as  in  the 
tecond  analysis.  If  he  is  at  a  loss  which  to  take  for  the  divisor,  let 
him  substitute  the  word  parts  for  the  denominator. 

3.  56  is  f  of  what?  7.  436  is  f  of  what? 

4.  68  is  |  of  what?  8.  456  is  £  of  what? 

5.  85  is  f  of  what  ?  9.  685  is  -^  of  what  ? 

6.  1  15  is  f  of  what  ?  10.  999  is  ^  of  what  ? 

1  1.  A  market  man  being  asked  how  many  eggs  he  had, 
replied  that  1  26  was  equal  to  -^  of  them  :  how  many  had  he  '* 

12.  If  -j^-  of  a  ship  is  worth  $8280,  what  is  the  whole 
worth  ? 

13.  A  commander  lost  f  of  his  forces  in  a  battle,  and 
had  9500  men  left:  how  many  had  he  at  first? 

14.  %%  is  \  of  what  number? 

ANALYSIS.—  fg  is  i  of  4  times  fg  ;  and  4  times  $£=2|. 

J5-  fi  is  f  °f  what  number  ? 

ANALYSIS.  —  Since  ii=$  of  a  certain  number,  \  of  that  number 
must  be'i  of  fi,  and  \  of  H=A»  or  \.  Now  if  |  of  the  number=i, 
$  must  equal  7  times  \=\,  or  i 


7,  21     *     7 

Or,  dividing  H  by  f  we  have       —  X  £=-,  or  i 
4,  ^o>     A     4 

1  6.  ||  is  |  of  what  number  ? 
17.  1%  is  f  of  what  number? 
1  8.  i8f  is  f  of  what  number? 

ANALYSIS.—  18}  =^5  .    Since  *4*=$,  i  =  14«,  and  t=*-l*,  or  30, 
Or,  ^-j-frr1^  x  g=»4o0Q=30,  the  same  as  before. 

J9-  3?i  is  f  of  what  number? 

20.  66f  is  f  of  what  number  ? 

21.  48  is  |  of  f  of  what  number  ? 

ANALYSIS.  —  f  of  J=f.    The  question  now  is,  48  is  f  of  what 
number?    Ana.  48  x  \,  or  108. 


138  FRACTIONAL     RELATIONS. 

22.  112  is  f  of  f  of  what  number? 

23.  In  -£  of  120  how  many  times  15  ? 

ANALYSIS.  —  £  °f  I2°  is  r3a  J  an(i  5  arc  7  times  13  i  or  93  ^.    Now 
=rY«a  or  6*,  Ana. 


24.  How  many  yards  of  brocatelle,  at  89  a  yard,  can  be 
bought  for  |  of  $100  ? 

25.  A  man  paid  -^  of  $280  for  84  arm-chairs:  what  was 
that  apiece  ? 

26.  90  is  £  of  how  many  times  17  ? 

ANALYSIS.  —  As  90  is  ?  of  a  certain  number,  1  is  ^  of  90,  which  is 
15  ;  and  ?  are  7  times  15  or  105.  Now  17  is  in  105,  6  -ft-  times. 
Therefore,  etc. 

27.  125  is  f  of  how  many  times  20  ? 

28.  A  man  paid  60  cents  for  his  lunch,  which  was  T6ff  of 
his  money,  and  spent  the  remainder  for  cigars,  which  were 
5  cents  each:  how  much  money  had  he;  and  how  many 
cigars  did  he  buy  ? 

29.  ^  of  1  10  is  ^  of  what  number  ? 

ANALYSIS.  —  ^  of  no  is  n,  and  ,9o,  9  times  n  or  99.  Now,  since 
99  is  ?  of  a  number,  |  of  it  must  be  ^  of  99,  which  is  n,  and  f  must 
be  7  times  n  or  77.  Therefore,  etc. 

30.  |  of  126  is  |  of  what  number  ? 

31.  |  of  90  is  |  of  how  many  times  1  1  ? 

ANALYSIS.  —  §  of  90  is  50.  Now  as  50  is  f  of  a  number,  £  is  |  of 
50  or  10  ;  |  is  8  times  10  or  80.  Finally,  n  is  contained  in  80,  7ft- 
times.  Therefore  §  of  90  is  §  of  7  ft-  times  n. 

32.  |  of  96  is  -^  of  how  many  times  20  ? 

33.  £  of  120  is  f  of  how  many  sevenths  of  56  ? 

ANALYSIS.  —  I  of  120  is  100.  If  100  is  J  of  a  number,  £  is  \  of  the 
number;  now  \  of  100  is  25,  and  §  5s  9  times  25  or  225.  Finally,  | 
of  56  is  8,  and  8  is  contained  in  225,  28^  times.  Therefore,  £  of  120 
is  J  of  28  i  times  |  of  56. 

34.  f  of  35  is  I  of  how  many  tenths  of  120  ? 


DECIMAL   FRACTIONS. 

175.  Decimal  Fractions  are  those  in  which  the 
unit  is  divided  into  tenths,  hundredths,  thousandths,  etc. 
.  They  arise  from  continued  divisions  by  10. 

If  a  unit  is  divided  into  ten  equal  parts,  the  parts  are 
called  tenths.  Now,  if  one  of  these  tenths  is  subdivided 
into  ten  other  equal  parts,  eacli  of  these  parts  will  be  one- 
tenth  of  a  tenth,  or  a  hundredth.  Thus,  -fo-7-  I0  or  fV  °f 
-I^=-IJir.  Again,  if  one  of  these  himdredths  is  subdivided 
into  ten  equal  parts,  each  of  these  parts  will  be  one-tenth  of 
a  hundredth,  or  a  thousandth.  Thus,  -j^u-r  io=rJ<rs>  etc« 


NOTATION    OF    DECIMALS. 

176.  If  we  multiply  the  unit  i  by  10  continually,  it 
produces  a  series  of  whole  numbers  which  increase  regu- 
larly by  the  scale  of  10  ;  as, 

I,    10,    100,    1000,    i  oooo,    I  ooooo,    i  oooooo,    etc. 

Now  if  we  divide  the  highest  term  in  this  series  by  10 
continually,  the  several  quotients  will  form  an  inverted 
series,  which  decreases  regularly  by  ten,  and  extends  from 
the  highest  term  to  i,  and  from  i  to  Js,  T^,  T^OTT*  and  so 
on,  indefinitely  ;  as, 

i  ooooo,   i  oooo,  1000,  100,  10,  i,  -j^-,  -j^,  T^OTT,   etc. 


177.  By  inspecting  this  series,  the  learner  will  perceive 
that  the  fractions  thus  obtained,  regularly  decrease  toward 
the  right  by  the  scale  of  10. 

If  we  apply  to  this  class  of  fractions  the  great  law  of 
Arabic  Notation,  which  assigns  different  values  to  figures, 

175.  What  are  decimal  fractions  ?  How  do  they  arise  ?  Explain  this  upon 
the  blackboard.  177.  By  what  law  do  decimals  decrease  ? 


.140  DECIMAL     FRACTIONS. 

according  to  the  place  they  occupy,  it  follows  that  a  figure 
standing  in  the  first  place  on  the  right  of  units,  denotes 
tenths,  or  i  tenth  as  much  as  when  it  stands  in  units' 
place ;  when  standing  in  the  second  place,  it  denotes  hun- 
dredths,  or  i  tenth  as  much  as  in  the  first  place;  when 
standing  in  the  third  place,  it  denotes  thousandths,  etc., 
each  succeeding  order  below  units  being  one  tenth  the 
value  of  the  preceding.  Hence, 

178.  Fractions  which  decrease  by  the  scale  of  ten,  may 
be  expressed  like  whole  numbers ;  the  value  of  each  figure 
in  the  decreasing  scale  being  determined  by  the  place  it 
occupies  on  the  right  of  units.  Thus,  3  and  5  tenths  may 
be  expressed  by  3.5 ;  3  and  5  hundredths  by  3.05 ;  3  and 
5  thousandths  by  3.005,  etc. 

178,  «•  Decimals  are  distinguished  from  whole  numbers 
by  a  decimal  point. 

NOTES. — i.  The  decimal  point  commonly  used  (.),  is  a  period. 
2.  This  class  of  fractions  is  called  decimals,  from  the  Latin  decem, 
ten,  which  indicates  both  their  origin  and  the  scale  of  decrease. 

179.  The  Denominator  of  a  decimal  fraction  is 
always    10,  100,  1000,  etc.;    or  i  with  as  many  ciphers 
annexed  to  it  as  there  are  decimal  places  in  the  given 
numerator.     Conversely, 

The  Numerator  of  a  decimal  fraction,  when  written 
alone,  contains  as  many  figures  as  there  are  ciphers  in  the 
denominator.  Thus  ^,  -^  T^nfJ  expressed  decimally, 
are  .5,  .05,  .005,  etc.  If  the  ciphers  in  .05  and  .005  are 
omitted,  each  becomes  5-tenths. 

What  place  do  tenths  occupy  ?  Hundredths,  thousandths,  etc.  f  178.  How  is 
the  value  of  decimal  figures  determined  ?  178,  a.  How  are  decimals  distinguished 
from  whole  numbers  ?  Note.  What  is  the  decimal  point  ?  From  what  does  this 
rlass  of  fractions  receive  its  name  ?  179.  What  is  the  denominator?  How  do 
the  number  of  ciphers  in  the  denominator  and  the  decimal  places  in  the  numera- 
tor compare  ? 


DECIMAL     FRACTIONS.  141 

180.  The  different  orders  of  decimals,  and  their  relative 
position,  may  be  seen  from  the  following 


TABLE. 


•2  =* 

;§  .a 

rf     a  -g 


fl 

>s 

§ 

tr. 

o 

^ 

2 

00 

.a 

| 

fl 

g 

o 

rfl 

«=' 

1 

1 

"S 

"3 
5 

OD 

~ 

••5 

s 

T3 

g 

p 

o 

— 

1 
tJ 

"s 
•2 

E3 

a 

••o 
£ 
^ 

1 

1 

Q 

a 

9) 

EH 

B 

1 

a 
6 
EH 

- 
0 

B 

1 

d 

® 

EH 

§ 

B 

1 

• 

3 

2 

6 

7 

2 

5 

4 

5 

8 

W         X       ^        -«         03 

*~  'S    C<    s    IS 

ill    i  1  s 

^»     ^       J    ^»     ri 

^     K      EH     ^i     B      EH 

842567 

V  

Integers.  Decimals. 

NOTES. — i.  A  Decimal  and  an  Integer  written  together,  are  called 
ft  Mixed  Number ;  as  35.263,  etc.  (Art.  101,  Def.  12.) 

2.  A  Decimal  and  a  Common  Fraction  written  together,  are  called 
a  Mixed  Fraction  ;  as  .6^,  .33 $. 

181.  Since  the  orders  of  decimals  decrease  from  left  to 
right  by  the  scale  of  i  o,  it  follows : 

First.  Prefixing  a  cipher  to  a  decimal  diminishes  its  value 
io  times,  or  divides  it  by  io.  Thus  .7=-^;  •°7=T0Lff> 


Annexing  a  cipher  to  a  decimal  does  not  alter  its  value. 
Thus  .7,  .70,  .700  are  respectively  equal  to  T 


These  effects  are  the  reverse  of  those  produced  by  annexing  and 
prefixing  ciphers  to  whole  numbers.  (Arts.  57.  79.) 

Second.  Each  removal  of  the  decimal  point  one  figure 
io  the  right,  increases  the  number  10  times,  or  multiplies 
it  by  io.  Each  removal  of  the  decimal  point  one  figure 
to  the  left,  diminishes  the  number  io  times,  or  divides  it 
by  io.  Hence. 

180.  Name  the  orders  of  decimals  toward  the  right.  Name  the  orders  of 
Integers  toward  the  left.  181.  How  is  a  decimal  figure  affected  by  moving  it  one 
place  to  the  right?  How,  if  a  cipher  is  prefixed  to  it?  How  if  ciphers  are 
annexed  ? 


142  DECIMAL     FRACTIONS. 

182.  To  write  Decimals. 

RULE. —  Write  the  figures  of  the  numerator  in  their  order, 
assigning  to  each  its  proper  place  below  units,  and  prefix  to 
them  the  decimal  point. 

If  the  numerator  has  not  as  many  figures  as  required, 
supply  the  deficiency  by  prefixing  ciphers. 

"Write  the  following  fractions  decimally : 

!•  rV  6- 

2-  iVff-  7- 

3.  T^T.  8.     4TV  13-   i0ioso°o8oo- 

4-  rVcr-  9-  2i-dhy  14- 

5.  yfo.  10.  84T4SV  15. 

17.  Write   6  hundredths ;    63  thousandths;    109  ten 
thousandths. 

1 8.  Write  305  thousandths;  21  hundred-thousandths; 
95  millionths. 

19.  Write  4  thousandths ;  1 08  ten-thousandths;  46  hun- 
dreths;  65  millionths;  1045  ten-millionths. 

20.  Write  sixty-nine  and  four  thousandths;  ten  and 
seventy-five  ten -thousands;  160  and  6  millionths. 

21.  Write   53  ten-thousandths;   63   and   28  hundred- 
thousandths;  352  ten-millionths. 

183.  To  read  Decimals  expressed  by  Figures. 

RULE. — Read  the  decimals  as  whole  numbers,  and  apply 
to  them  the  name  of  the  lowest  order. 

NOTE. — In  case  of  mixed  numbers,  read  the  integral  part  as  if  it 
Btood  alone,  then  the  decimal. 

Or,  pronounce  the  word  decimal,  Jhen  read  the  decimal  figures  aa 
if  they  were  whole  numbers. 

182.  What  is  the  rnle  for  writing  decimals?  Note.  If  the  numerator  has  not 
as  many  figures  as  there  are  ciphers  in  the  denominator,  what  is  to  be  done? 
183.  How  are  decimals  read  ?  Note.  In  case  of  a  mixed  number,  how  ? 


DECIMAL     FRACTIONS.  143 

Or,  having  pronounced  the  word  decimal,  repeat  the  names  of  the 
decimal  figures  in  their  order.  Thus,  275.468  is  read,  "  275  and  468 
thousandths;"  or  "  275,  decimal  four  hundred  and  sixty-eight;"  or 
"  275,  decimal  four,  six,  eight" 

i.  Explain  the  decimal  .05. 

ANALYSIS. — Since  the  5  stands  in  the  second  place  on  the  right  of 
the  decimal  point,  it  is  equivalent  to  TfSr>,  and  denotes  5  hundredth^ 
of  one,  or  5  such  parts  as  would  be  obtained  by  dividing  a  unit  into 
100  equal  parts. 

Read  the  following  examples : 

('•)  ('.)  (3-) 

.36  2.751  32.862 

•479  4-8465  40.0752 

.0652  7-25025  57.00624 

.00316  8.400452  81.20701 

183,  <(•  Decimals  differ  from  common  fractions  in  three 
respects,  viz.  :  in  their  origin,  their  notation,  and  their  limitation. 

ist.  Common  fractions  arise  from  dividing  a  unit  into  any  number 
of  equal  parts,  and  may  have  any  number  for  a  denominator. 

Decimals  arise  from  dividing  a  unit  into  ten,  one  hundred,  one 
thousand,  etc.,  equal  parts  ;  consequently,  the  denominator  is  always 
10,  or  some  power  of  10.  (Arts.  55,  n.  179.) 

2d.  In  the  notation  of  common  tractions,  both  the  numerator  and 
denominator  are  written  in  full. 

In  decimals,  the  numerator  only  is  written ;  the  denominator  is 
understood. 

3d.  Common  fractions  arc  universal  in  their  application,  embracing 
all  classes  of  fractional  quantities  from  a  unit  to  an  infinitesimal. 

Decimals  are  limited  to  that  particular  class  of  fractional  quanti- 
ties whosa  orders  regularly  decrease  in  value  from  left  to  right,  by 
the  scale  of  10. 

NOTE. — The  question  is  often  asked  whether  the  expressions  -|a0-, 
T^U»  TiiVo.  etc.,  are  common  or  decimal  fractions. 

All  fractions  whose  denominator  is  written  under  the  numerator, 
fulfil  the  conditions  of  common  fractions,  and  may  be  treated  as  such. 
But  fractions  which  arise  from  dividing  a  unit  into  10, 100,  1000,  eta, 
equal  parts,  answer  to  the  definition  of  decimals,  whether  the  denom- 
inator is  expressed,  or  understood. 

183,  a.  llow  do  decimals  differ  from  common  fraction.- ? 


144  DECIMAL    FBACTIONS. 

REDUCTION   OF   DECIMALS. 
184.  To  Reduce  Decimals  to  a  Common  Denominator, 

1.  Eeduce  .06,  2.3,  and  .007  to  a  common  denominator. 
ANALYSIS  —  Decimals  containing  the  same  number       .06  —  0.060 

of  figures,  have  a  common  denominator.    (Art.  179.)       2  .?  —  2  ,oo 
By  annexing  ciphers,  the  number  ot  decimal  fig-  _ 

ares  in  each  may  be  made  the  same,  without  altering 
their  value.    (Art.  181.)    Hence,  the 

RULE.  —  Make  the  number  of  decimal  figures  the  same  in 
each,  by  annexing  ciphers.     (Art.  181.) 

2.  Reduce  .48  and  .0003   to  a  common  denominator. 

3.  Reduce  2  to  tenths  ;    3  to  hundredths  ;   and  .5  to 
thousandths.  Ans.  2.0  or  f£;  3.00  or  f£&;  .500. 


185.  To  Reduce  Decimals  to  Common  Fractions. 

1.  Reduce  .42  to  a  common  fraction. 

ANALYSIS.  —  The  denominator  of  a  decimal  is  i,  with  as  many 
ciphers  annexed  as  there  are  figures  in  its  numerator;  therefore  the 
denominator  of  .42  is  too.  (Art.  179.)  Ans,  .42=-,^-.  Hence,  the 

RULE.  —  Erase  the  decimal  point,  and  place  the  denomi- 
nator lender  the  numerator.  (Art.  179.) 

2.  Reduce  .65  to  a  common  fraction  ;  then  to  its  lowest 
terms.  Ans.  .65  =£&,  and  -r^=i£- 

Reduce  the  following  decimals  to  common  fractions  : 

3.  .128  7.  .05  n.  .0007          15.  .200684 

4.  .256  8.  .003  12.  .04056        16.  .0000008 

5.  .375  9.  .0008  13.  .00364        17.  .12400625 

6.  .863  10.  .0605  14.  .00005        18.  .24801264 

186.  To  Reduce  Common  Fractions  to  Decimals. 
i.  Reduce  f  to  a  decimal  fraction. 

ANALYSIS  —  \  equals  £  of  3.   Since  3  cannot  be  divided  OPERATION 

by  8;  we  annex  a  cipher  to  reduce  it  to  tenths.    Now}  8)3.000 

0/30  tenths  is  3  tenths,  and  6  tenths   over.     6  tenths  Ans     -Z7C 
reauced   to  hundredths  =60  hundredths,  and    J  of  60 
hundredths=7  hundredths  and  4  hundredths  over.     4  hundredth* 

reduced  to  thousandths  =40  thousandths,  and  i  of  40  thousandths 
=5  thousandths.    Therefore,  £=.375.    Hence,  the 


DECIMAL     FRACTIONS  145 

RULE — Annex  ciphers  to  the  numerator  and  divide  by  the 
denominator.  Finally,  point  off  as  many  decimal  figures  in 
the  result  as  there  are  ciphers  annexed  to  the  numerator. 

NOTE.— If  the  number  of  figures  in  the  quotient  is  less  than  tha 
number  of  ciphers  annexed  to  the  numerator,  supply  the  deficiency 
by  prefixing  ciphers. 

Demonstration  — A  fraction  indicates  division,  and  its  value  is  the 
numerator  divided  by  the  denominator.  (Arts.  134,  142.)  Now,  an- 
nexing one  cipher  to  the  numerator  multiplies  the  fraction  by  10 ; 
annexing  two  ciphers,  by  100,  etc.  Hence,  dividing  the  numerator 
with  one,  two,  or  more  ciphers  annexed,  gives  a  quotient,  10,  roo. 
etc.,  times  too  large.  To  correct  this  error  the  quotient  is  dividad 
by  10,  100,  looo,  etc.  But  dividing  by  10,  100,  etc.,  is  the  same  as 
pointing  off  an  equal  number  of  decimal  figures.  (Art.  181.) 

Reduce  the  following  fractions  to  decimals : 

2.  i  6.  |  10.  $£  14.  .g-^ 

3-  f  7-  fV  "•  s3S  J5-  -jfzr 

4-  I  3.  -I  12.  ^  1 6.  jd-s 

5-  t  9-  -h3  J3-  TTO-  17-  xMir 

1 8.  Reduce  f  to  the  form  of  a  decimal. 

ANALYSTS. — Annexing  ciphers  to  the  numerator      OPERATION. 
and  dividing  by  the  denominator,  as  before,  the  quo-    3)2.000 
tient  consists  of  6  repeated  to  infinity,  and  the  re-        .6666   etc, 
maiuder  is  always  2.    Therefore  §  cannot  be  exactly 
expressed  by  decimals. 

19.  Reduce  75T  to  the  form  of  a  decimal. 

ANAi/ys'B. — Having  obtained  th~ee  quotient  OPERATION. 

figures  135,  the  remainder  is  5,  the  saoie  as  the    37(5.000000 
original  numerator;   consequently,  by  annex-  .  I^I  "55    etc. 

ing  ciphers  to  it,  and  continuing  the  division, 
we  obtain  the  same  set  of  figures  as  before,  repeated  to  infinity. 
Therefore  ^  cannot  be  exactly  expressed  by  decimals. 

187.  When  the  numerator,  with  ciphers  annexed,  is 
exactly  divisible  by  the  denominator,  the  decimal  is  called 
a  terminate  decimal. 

185.  How  reduce  a  decimal  to  a  common  fraction  ?    186.  How  reduce  a  com- 
mon fraction  to  a  decimal  ?    Note.  If  th<?  number  of  flares  in  the  quotient  i 
t'lan  that  in  the  numerator,  what  is  to  be  done  ?   Explain  the  reason  for  pointing 
off  the  quotient. 


146  ADDITION     OF     DECIMALS. 

When  ifc  is  not  exactly  divisible,  and  the  same  figure  a 
set  of  figures  continually  recurs  in  the  quotient,  the  deci 
mal  is  called  an  interminatc  or  cii  culatiny  decimal. 

The  figure  or  set  of  figures  repeated  is  called  the  repetend 
Thus,  the  decimals  obtained  in  the  last  two  examples  ar< 
interminate,  because  the  division,  if  continued  forever,  will 
leave  a  remainder.  The  repetend  of  the  i8th  is  6;  thai 
of  the  i  gth  is  135. 

NOTES. — i.  After  the  quotient  has  been  carried  as  far  as  desirably 
the  sign  (  +  )  is  annexed  to  it  to  indicate  there  is  still  a  remainder. 

2.  If  the  remainder  is  such  that  the  next  quotient  figure  would  be 
5,  or  more,  the  last  figure  obtained  is  sometimes  increased  by  i,  and 
the  sign  (— )  annexed  to  show  that  the  decimal  is  too  large. 

3.  Again,  the  remainder  is  sometimes  placed  over  the  divixor  and 
annexed  to  the  quotient,  forming  a  mixf.d  fraction.    (Art.  i8c,  n.) 
Thus  if  jj-  is  reduced  to  the  decimal  form,  the  result  may  be  expressed 
by  .6666+  ;  by  .6667—  ;  or  by  .6666  f. 

(For  the  further  consideration  of  Circulating  Decimals,  the  student 
is  referred  to  Higher  Arithmetic.) 

Reduce  the  following  to  four  decimal  places : 
20.    $  22.     f  24.    T\  26.    £f 

2L      I  23.      A  25-      •&  27.      ft 

Reduce  the  following  to  the  decimal  form : 

28.  75$      30.  261^      32.  465^0      34-  74°*VV 

29.  i36|      31.  346^4.      33.  523^      35.  956^ 

ADDITION    OF    DECIMALS. 

188.  Since  decimals  increase  and  decrease  regularly  by 
the  scale  of  10,  it  is  plain  they  may  be  added,  subtracted, 
multiplied,  and  divided  like  whole  numbers. 

Or,  they  may  be  reduced  to  a  common  denominator,  then 
be  added,  subtracted,  and  divided  like  Common  Fractions. 
(Arts.  156,  184.) 

187.  When  the  numerator  with  ciphers  annexed  is  exactly  divisible  by  the 
denominator,  what  SB  the  decimal  called  ?  If  not  exactiv  divisible,  what  ?  What 
is  the  figure  or  set  of  figures  repeated  called? 


ADDITION    OF    DECIMALS.  147 

189.  To  find  the  Amount  of  two  OP  more  Decimals. 

i.  Add  360.1252,  1.91,  12.643,  and  152.8413. 

ANALYSIS. — Since   units   of  the  same  order  or  OPERATION. 

like  numbers  ouiv  can  be  added  to  each  other,  we  360.1252 

reduce  the  decimals  to  a  common  denominator  by  1 .9 1  oo 

annexing  ciphers;    or,   which  is   the   same,  by  12.6430 

writing  the  decimals  one  under  another,  so  that  152.8413 

the  decimal  points  shall  be  in  a  perpendicular  Ans.  $27.5,  los 
line.  (Arts.  28,  n.  156.)  Beginning  at  the  ri jhi, 
we  add  each  column,  and  set  down  the  result  as  in  whole  numbers, 
and  for  the  same  reasons.  (Art.  29,  n.)  Finally,  we  place  the  deci- 
mal point  in  the  amount  directly  under  those  in  the  numbers  added. 
(Art.  178,  a)  Hence,  the 

RULE. — I.  Write  the  numbers  so  that  the  decimal  points 
shall  stand  one  under  another,  with  tenths  under  tenths,  etc. 

II.  Beginning  at  the  right,  add  as  in  whole  numbers,  and 
place  the  decimal  point  in  the  amount  under  those  in  the 
numbers  added. 

NOTE. — Placing  tenths  under  tenths,  hundredths  under  hundredths, 
etc.,  in  effect,  reduces  the  decimals  to  a  com.  denominator;  hence 
the  ciphers  on  the  right  may  be  omitted.  (Arts.  181,  184.) 

('•)  (3)  (4-)  (5.) 

41.3602  416.378  36.81045  4-83907 

4.213  85.1  .203  .293 

61.46  .4681  5-3078  .40 

375-265  4.38  87.69043  5.1067 

482.2982  Ans.  375-2956  9.25  3.75039 

6.  AVI uit  is  the  sum  of  41.37 1  +  2.29  +  73.402  +  1.729  ? 

7.  What  is  the  sum  of  823.37  +  7.375  +  61.1 +.843  ? 

8.  What  is  the  sum  of  .3925 +  .64  +  .462  +  . 7 +.56781  ? 

9.  AVhat  is  the  sum  of  86  005+4.0003  +  2.00007  ? 
to.  AVhat  is  the  sum  of  1.713  +  2.30  +  6.400+27.004? 

11.  Add  together  7  tenths;  312  thousandths;  46  hun« 
credths;  9  tenths;  and  228  ten-thousandths. 

12.  Add  together   23   ten-thousandths;    23   hundred- 
thousandths;  23  thousandths;  23  hundredths;  and  23. 

1^9.  How  are  decimals  added  f    How  point  off  the  amount  ? 


\ 

148  SUBTRACTION     OF     DECIMALS. 

13.  Add  together  five  hundred  seventy-five  and  seven- 
tenths;   two   hundred  fifty-nine   ten-thousandths;   five- 
millionths;  three  hundred  twenty  hundred-thousandths. 

14.  A  farmer  gathered  iy|  bushels  of  apples  from  one 
tree ;  8^  bushels  from  another;  10^  bushels  from  another ; 
and  i6|  bushels  from  another.     Required  the  number  of 
bushels  he  had,  expressed  decimally. 

15.  A  grocer  sold  7-^  pounds  of  sugar  to  one  customer; 
11.37  pounds  to  another;    icf  pounds  to  another;  25^ 
pounds  to  another;  and  21.375  pounds  to  another:  how 
many  pounds  did  he  sell  to  all  ? 

SUBTRACTION   OF   DECIMALS. 

190.  To  find  the  Difference  between  two  Decimals. 

j.  What  is  the  difference  between  2.607  an(l  -7235  ? 

ANALYSIS. — We  red  uce  the  decimals  to  a  common  OPEKATIOH. 

denominator,  by  annexing  ciphers,  or  by  writing  the  2.607 

game  orders  one  under  another.     (Art.  189,  n.)    For,  *7235 

units  of  the  game  order  or  like  numbers  only,  can     Ans.  1.8835 
be  subtracted  one  from  the  other. 

Beginning  at  the  right,  we  perceive  that  5  ten-thousandths  can 
not  be  taken  from  o ;  we  therefore  borrow  ten,  and  then  proceed  in 
all  respects  as  in  whole  numbers.  (Art.  38.)  Hence,  the 

RULE. — I.  Write  the  less  number  tender  the  greater}  so 
that  the  decimal  points  shall  stand  one  under  the  oilier, 
with  tenths  under  tenths,  etc. 

II.  Beginning  at  the  right  hand,  proceed  as  in  subtract- 
ing whole  numbers,  and  place  the  decimal  point  in  the 
remainder  tinder  that  in  the  subtrahend. 

NOTE. — Writing  the  same  orders  one  under  another,  in  effect 
reduces  the  decimals  to  a  common  denominator.  (Art.  189,  n.) 

(*•)  (3-)  (4.)  (5-) 

From  13.051        7-0392         20.41         85.3004 

Take     5.22  .4367 1  3-°42S  67-35246 


190.  How  are  decimals  subtracted  ?    How  point  off  the  remainder  T 


MULTIPLICATION    OFDECIMALS.  149 

Perform  the  following  subtractions : 

6.  13.051  minus  5.22.  12.  10  minus  9.1030245. 

7.  7.0392  minus  0.43671.         13.  100  minus  99.4503067. 

8.  20.41  minus  3.0425.  14.  i  minus  .123456789. 

9.  85.3004  minus  7.35246.       15.  i  minus  .98764321. 

10.  93.38  minus  14.810034.       16.  o.i  minus  .001. 

11.  3  minus  0.103784.  17.  o.oi  minus  .oooor. 

18.  From  100  take  i  thousandth. 

19.  From  45  take  45  ten-thousandths. 

20.  From  i  ten-thousandth  take  2  millionths. 

21.  A  man  having  $673.875,  paid  $230.05  for  a  horse: 
how  much  had  he  left  ? 

22.  A  father  having  504.03  acres  of  land,  gave   100.45 
acres  to  one  son,  263.75  acres  to  another:  how  much  had 
he  left  ? 

23.  What  is  the  difference  between  203.007  and  302.07  ? 

24.  Two  men  starting  from  the  same  place,  traveled  in 
opposite  directions,  one  going  571.37   miles,   the  other 
501.037  miles:  how  much  further  did  one  travel  than  the> 
other ;   and  how  far  apart  were  they  ? 

MULTIPLICATION   OF   DECIMALS. 
191.  To  find  the  Product  of  two  or  more  Decimals. 

1.  What  is  the  product  of  45  multiplied  by  .7  ? 

ANALYSIS. — -T^^s-  Now -^-times  45  =  Vcf  =315-5-10,  OPERATION. 
or  31  5,  Ana.    In  the  operation,  we  multiply  by  7  in-         45 
stead  of  -fo  ;  therefore  the  produc*  is  10  times  too  large.          .7 

To  correct  this,  we  point  off  i   figure   on   the   right,  -II.G  Ant. 
which  divides  it  by  10.    (Arts.  79,  143,  165.) 

2.  What  is  the  product  of  9.7  multiplied  by  .9  ? 
ANALYSIS. — 9.7=?$,  and  .9=-,V    Now  -f{J  times  ?£=    OPKRATIOK. 

^1=873-:- loo,  or   8.73,  Ans.     In  this  operation   we  9-7 

also  multiply  as  in   whole  numbers,  and   point  off  2  .9 

figures  on  the  right  of  the  product  for  decimals,  which  8.73 
divides  it  by  100.    (Art.  79.) 


150  MULTIPLICATION    OF    DECIMALS. 

BEM. — By  inspecting  these  operations,  we  see  that  each  answer 
contains  as  many  decimal  figures  as  there  are- decimal  places  in  both 
factors.  Hence,  the 

KULE. — Multiply  as  in  whole  numbers,  and  from  the 
right  of  the  product,  point  off  as  many  figures  for  decimals, 
as  there  are  decimal  places  in  both  factors. 

NOTES. — i.  Multiplication  of  Decimals  is  based  upon  the  eame 
principles  as  Multiplication  of  Common  Fractions.  (Art.  166.) 

2.  The  reason  for  pointing  off  the  product  is  this:  The  product  of 
any  two  decimal  numerators  is  as  many  times  too  large  as  there  arc 
units  in  the  product  of  their  denominators,  and  pointing  it  off  divides 
it  by  that  product.  (Art.  79.)  For,  the  product  of  the  denominators 
of  two  decimals  is  always  i  with  as  many  ciphers  annexed  as  there 
are  decimal  places  in  both  numerators.  (Arts.  57,  179,  186,  Dem.) 
.  3.  If  the  produ,  has  not  as  many  figures  as  there  are  decimals  in 
both  factors,  supply  the  deficiency  by  prefixing  ciphers. 

2.  Multiply  .015  by  .03. 

SOLUTION. — .015  x  ,03=.ooo45.  The  product  requires  5  decimal 
places ;  hence,  3  ciphers  must  be  prefixed  to  45. 

(3-)  (4-)  (5-)  (60 

Multiply         29.06  .07213  .000456  4360.12 

By  .005  .OO2I  -0037  5.000 

7.  4.0005  x  .00301.  ii.  0.0048  x  .0091. 

8.  5.0206x4.0007.  12.   15.004  x  .10009. 

9.  3.0004x106.  13.  6.0103  x  .00012. 
10.  7.2136x100.  14.  20007  x  .000001. 

15.  If  i  box  contains  17.25  pounds  of  butter,  how  many 
pounds  will  25  boxes  contain  ? 

1 6.  What  cost  20. <;  barrels  of  flour,  at  $10.875  a  barrel  ? 

17.  If  one  acre  produces  750.5  bushels  of  potatoes,  how 
much  will  .625  acres  produce  ? 

18.  What  cost  53  horses,  at  $200.75  apiece? 

19.  Multiply  28  hundredths  by  45  thousandths. 

20.  What  cost  73.25  yards  of  cJoth,  at  $9.375  per  yard  ? 

191.  How  are  decimals  multiplied  ?  How  point  off  the  product  ?  Note.  Ex- 
plain the  reason  for  pointing  off.  If  the  product  doea  not  contain  as  many 
figure?  as  there  are  dccimnl^  in  the  factors,  what  is  to  be  done  ?  192.  How  mul- 
tiply \  decimal  by  10,  100,  etc. 


MULTIPLICATION     OF     DECIMALS.          151 

21.  Multiply  5  tenths  by  5  thousandths. 

22.  Multiply  two  hundredths  by  two  ten-thousandths. 

23.  Multiply  seven  hundredths  by  seven  millionths. 

2  4..  Multiply  two  hundred  and  one  thousandths  by  three 
millionths. 

25.  Multiply  five  hundred-  thousandths  by  six  thous- 
andths. 

26.  Multiply  four  millionths  by  sixty-three  thousandths. 

27.  Multiply  a  hundred  by  a  hundred-thousandth. 

28.  Multiply  one  million  by  one  millionth. 

29.  Multiply  one  millionth  by  one  billionth. 

192.  To  multiply  Decimals  by  10,  100,  1000,  etc. 

30.  Multiply  .43215  by  1000. 


.  —  Removing  a  figure  one  place  to  the  left,  we  have  seen 
multiplies  it  by  10.  But  moving  the  decimal  point  one  place  to  the 
right  has  the  same  effect  on  the  position  of  the  figures  ;  therefore,  it 
multiplies  them  by  10.  For  the  same  reason,  moving  the  decimal 
point  two  places  to  the  right  multiplies  the  figures  by  100,  and  so  on. 
In  the  given  example  the  multiplier  is  1000  ;  we  therefore  move  the 
decimal  point  three  places  to  the  right,  and  have  432.15,  the  answer 
required.  Hence,  the 

RULE.  —  Move  the  decimal  point  as  many  places  toward 
the  right  as  there  are  ciphers  in  the  multiplier.  (Art.  181.) 

31.  Multiply  32.0505  by  ico. 

32.  Multiply  8.00356  by  1000. 

33.  Multiply  0.000243  by  10000. 

34.  Multiply  0.000058  by  100000. 

35.  Multiply  0.000005  by  1000000. 

36.  If  a  newsboy  makes  $0.005  on  each  paper,  what  is 
Irs  profit  on  10000  papers? 

37.  What  is  the  profit  on  100000  eggs,  at  $0.006  apiece? 

38.  If  a  farmer  gives  4.25  bushels  of  apples  for  one  yard 
of  cloth,  how  many  bushels  should  he  give  for  6.5  yards  ? 

39.  If  a  man  walks  3.75  miles  an  hour,  how  far  will  he 
walk  in  17.5  hours? 


152  DIVISION    OF     DECIMALS. 

DIVISION   OF    DECIMALS. 
193.   To  divide  one  Decimal  by  another. 

1.  How  many  times  .2  in  .3  ? 

ANALYSIS.  —  These    decimals    have    a    com.  de-         .2).8 
nom.  ;  hence,  we  divide  as  in  Common  Fractions,      Anv~Al\m  &. 
and  the  quotient  is  a  whole  number.    (Art.  169,  n.) 

2.  How  many  times  .08  in  .7  ? 

ANALYSIS.  —  Reduced  to  a  common  denominator,  the      .o8).7ooo 
given  decimals  become  .08  and  .70.     Now  .08  is  in  .70,       A        g 
8  times,  and  .06  rem.     Put  the  8  in  units'  place.     Re- 
duced to  the  next  lower  order,  .06=.  060,  and  .08  is  in  .060,  .7  of  a 
time,  and.  004  rem.   Put  the  7  in  tentlis'  place.   Finally,  .08  is  in  .0040, 
.05  of  a  time.    Write  the  5  in  hundredth*'  place.    Ana.  8.75  times. 

3.  How  many  times  is  .5  contained  in  .025  ? 
ANALYSIS.  —  Reduced  to  a  common  denominatoi,     ,5oo).O25oo 

.5=.  500,  and  .O25=.o25.     Since  500  is  not  contained 


..       ,          ...  „ 

•    -i  •  IT  -/l/fco»  O.Os  i. 

in  .025,  put  a  cipher  in  units  place,  ana  reduce  to 
the  next  lower  order.     But  .500  is  not  contained  in  .0250.     Put  a 
cipher  in  tenths'  place,  and  reducing  to  the  next  order,  .500  is  in 
.02500,  .05  of  a  time.    Write  the  5  in  hundredths'  place. 

REM.  —  When  two  decimals  have  a  com.  denom.  the  quotient  figures 
thence  arising  are  whole  numbers,  as  in  common  fractions.  (Art.  169,  n.) 

If  ciphers  are  annexed  to  the  remainder,  the  next  quotient  figure 
will  be  tenths,  the  second  hundredths,  &c.  Hence,  the 

EULW  —  Reduce  the  decimals  to  a  common  denominator, 
and  divide  the  numerator  of  the  dividend  by  that  of  the 
divisor,  placing  a  decimal  point  on  the  right  of  the  quotient. 

Annex  ciphers  to  the  remainder,  and  divide  as  before. 
The  figures  on  the  left  of  the  decimal  point  are  whole  num- 
bers j  those  on  the  right,  decimals. 

Or,  divide  as  in  whole  numbers,  and  from  the  right  of  the 
quotient,  point  off  as  many  figures  for  decimals  as  the  deci- 
mal places  in  the  dividend  exceed  those  in  the  divisor. 

NOTES.  —  i.  If  there  are  not  figures  enough  in  the  quotient  for  the 
decimals  required  by  the  second  method,  prefix  ciphers. 

193.  How  divide  decimals?  If  there  is  a  remainder?  Rem.  When  two  decimals 
have  a  com.  denom.,  what  is  the  quotient  ?  When  ciphers  are  annexed  to  tha 
remainder,  what  ?  When  the  ad  method  is  used,  how  point  off  T 


DIVISION    OF    DECIMALS.  153 

2.  If  there  is  a  remainder  after  the  required  number  of  decimals  is 
found,  annex  the  sign  +  to  the  quotient. 

2.  Divide  .063  by  9.  4.  Divide  642  by  1.07. 

3.  Divide  .856  by  .214.         5.  Divide  4.57  by  n. 

Perform  the  following  divisions : 

6.  78.4-^-2.6.  14.  .03753-^.00006, 

7.  8.45 -r  3.5.  15.    I2H-I.2. 

8.  1.262  —  9/7.  l6.    I.2-T-.I2. 

9.  .4625  — .65.  17.    .12-:- 12. 

10.  97.68—100.  18.  .00001-7-5. 

11.  6.75-:-iooo.  19.  .00005-:-.]'. 

12.  .576-^10000.  20.  .0003-7- .00000 6. 

13.  45.30-7-3020.  21.   .27H-IOOOOOO. 

22.  If  2.25  yards  of  cloth  make  i  coat,  how  many  coata 
can  be  made  of  103.5  yards  ? 

23.  How  many  rods  in  732.75  feet,at  16.5  feet  to  a  rod  ? 

24.  At  $18.75  apiece,  how  many  stoves  can  be  bought 
for  8506.25  ? 

194.  To  Divide  Decimals  by  10,  100,  1000,  etc. 

25.  "What  is  the  quotient  of  846.25  divided  by  100? 

ANALYSIS. — Moving  the  decimal  point  one  place  to  the  left  divides 
a  number  by  10.  For  the  same  reason,  moving  the  decimal  point 
two  places  to  the  left,  divides  it  by  100.  In  the  given  question,  re- 
j  loving  the  decimal  point  two  places  to  the  left,  we  have  8.4625,  the 
rnswer  required.  Hence,  the 

RULE. — Move  the  decimal  point  as  many  places  toward 
the  left  as  there  are  ciphers  In  the  divisor.  (Art.  181.) 

26.  Divide  4375.3  by  1000.         28.  Divide  2.53  by  100000. 

27.  Divide  638.45  by  10000.        29.  Divide  .5  by  1000000. 

30.  Bought  1000  pins  for  $.5 :  what  was  the  cost  of  each  ? 

31.  If  a  man  pays  $475  for  10000  yards  of  muslin,  what 
is  that  a  yard  ? 

194.  How  divide  decimals  by  10,  100,  1000,  etc.? 


195.  United  States  Money  is  the  national  cur- 
rency of  the  United  States,  and  is  often  called  Federal 
Money.    It  is  founded  upon  the  Decimal  Notation,  and  is 
thence  called  Decimal  Currency. 

Its  denominations  are  eagles,  dollars,  dimes,  cents,  and 
mills. 

TABLE. 

10    mills  (m.)        are  i  cent,  -    -    -    -    -    ft. 

10    cents  "    I  dime, d. 

10    dimes  "    i  dollar,     -    -  dol.  or  $. 

10    dollars  "   I  eagle,      -    -    -     -    E 

• 

NOTATION   OF   UNITED   STATES   MONEY. 

196.  The  Dollar  is  the   unit;   hence,   dollars   are 
whole  numbers,  and  have  the  sign  ($)  prefixed  to  them. 

In  $i  there  are  100  cents;  therefore  cents  are  liun- 
dredths  of  a  dollar,  and  occupy  hnndredths'  place. 

Again,  in  $i  there  are  1000  mills;  hence,  mills  are 
thousandths  of  a  dollar,  and  occupy  thousandths'  place. 

No  FES. — i.  The  origin  of  the  sign  ($)  has  been  variously  er- 
plaine  1.  Some  suppose  it  an  imitation  of  the  two  pillars  of  Her- 
cules, connected  by  a  scroll  found  on  the  old  Spanish  coins.  Others 
think  it  is  a  modified  figure  8,  stamped  upon  these  coins,  denoting 
8  reals,  or  a  dollar. 

A  more  plausible  explanation  is  that  it  is  a  monogram  of  United 
States,  the  curve  of  the  U  being  dropped,  and  the  S  written  over  it. 

195.  What  !s  United  States  money?  Upon  what  founded?  What  sometimes 
called?  The  denominations  ?  Repeat  the  Table.  196.  What  is  the  unit  ?  What  are 
dimes?  Cents?  Mills?  Note.  What  is  the  origin  of  the  sign  $ ?  The  meaning 
of  dime?  Cent?  Mill?  197.  How  write  United  States  money?  Note.  IIow  arc 
eagles  and  dimes  expressed?  If  the  number  of  cents  is  less  than  10,  what  mns»t 
be  done  ?  Why  ?  If  the  mill.'  are  5  or  more,  what  considered  ?  If  less,  what  ? 


UNITED     STATES     MONEY.  155 

2.  The  term  Dime  is  the  French  dixieme,  a  tentfi;  Cent  from  the 
Latin  centum,  a  hundred  ;  aud  Jtfi'W  from  the  Latin  mille,  a  thousand. 

3.  United  States  money  was  established  by  act  of  Congress,  in 
1786.     Previous  to  that,  pounds,  shillings,  pence,  etc.,  were  in  use. 

197.  To  express  United  States  money,  decimally. 
Ex.  i.  Let  it  be  required  to  write  75  dollars,  37  cents, 
and  5  mills,  decimally. 

ANALYSIS. — Dollars  are  integers;  we  therefore  write  the  75  dol- 
lars as  a  whole  number,  prefixing  the  ($),  as  $75.  Again,  cents  are 
hundredths  of  a  dollar ;  therefore  we  write  the  37  cents  in  the  first 
two  places  on  the  right  of  tho  dollars,  with  a  decimal  point  on  their 
left  as  $75.37.  Finally,  mills  are  thousandths  of  a  dollar ;  and  writing 
the  5  mills  in  the  first  place  on  the  right  of  cents,  we  have  $75.375, 
the  decimal  required.  (Art.  179.)  Hence,  the 

RULE. —  Write  dollars  as  ivhole  numbers,  cents  as  hun- 
dredths,  and  mills  as  thousandths,  placing  the  sign  ($)  be- 
fore dollars,  and  a  decimal  point  between  dollars  and  cents. 

NOTES. — I.  Eagles  and  dimes  are  not  used  in  business  calcula- 
tions ;  the  former  are  expressed  by  lens  of  dollars  ;  the  latter  by  tens 
of  cents.  Thus,  15  eagles  are  $150,  and  6  dimes  are  60  cents. 

2.  As  cents  occupy  two  places,  if  the  number  to  be  expressed  is  less 
than  10,  a  cipher  must  be  prefixed  to  the  figure  denoting  them. 

3.  Cents  are  often  expressed  by  a  common  fraction  having  100  for 
its  denominator.     Thus,  $7.38  is  written  $7  iVfl.  and  i8  rc^d  "  7  and 
-,3,,8(T  dollars." 

Mills  also  are  sometimes  expressed  by  a  common  fraction.  Thus, 
12  cts.  and  5  mills  are  written  $0.125,  or  $0.12^ ;  18  cts.  and  7^  mills 
are  written  $0.1875,  or  $o.i8J,  etc. 

4.  In  biisiness  calculations,  if  the  mills  in  the  result  are  5  or  more^ 
they  are  considered  a  cent ;  if  less  than  5,  they  are  omitted. 

1.  Write  forty  dollars  and  forty  cents. 

2.  "Write  five  dollars,  five  cents  and  five  mills. 

3.  Write  fifty  dollars,  sixty  cents,  and  three  mills. 

4.  Write  one  hundred  dollars,  seven  cents,  five  mills. 

5.  Write  two  thousand  and  one  dollars,  eight  and  a  half 
cents. 

6.  Write  7  hundred  and  5  dollars  and  one  cent. 

7.  Write  84  dollars  and  12^  cents. 

8.  Write  5  and  a  half  cents;  6  and  a  fourth  cents;  n 
and  three-fourths  cents,  decimally. 


15tf  UNITED     STATES     MOXEY. 

9.  Write  7  dollars,  31  and  a  fourth  cents,  decimally. 

10.  Write  19  dollars,  31 J  cents,  decimally. 

11.  Write  14  eagles  and  8  dimes,  decimally. 

12.  Write  5  eagles,  5  dollars,  5  cents,  and  5  mills. 

198.  To  read  United  States  money,  expressed  decimally. 
RULE. — Call  the  figures  on  the  left  of  the  decimal  point,dol- 

lars  ;  those  in  the  first  two  places  on  the  r  ight, cents  ;  the  next 
figure,  mills  ;  the  others,  decimals  of  a  mill. 

The  expression  $37.52748  is  read  37  dollars, '52  cts.,  7 
mills,  and  48  hundredths  of  a  mill. 

NOTE. — Cents  and  mills  are  sometimes  read  as  decimals  of  a  do?lar. 
Thus,  $7.225  may  be  read  7  and  225  thousandths  dollars. 

Read  the  following: 

1.  $204.30  5.    $78.104  9.  $1100.001 

2.  $360.05  6.     890.007  10.  $7.3615 

3.  §500.19  7.  $1001.10  ii.  $8.0043 

4.  $61.035  8.  $1010.01  12.  $10.00175 

REDUCTION  OF   UNITED    STATES   MONEY. 
CASE   I. 

199.  To  reduce  Dollars  to  Cents  and  Mills. 

1.  In  $67  how  many  cents? 

ANALYSIS. — As  there  arc  zoo  cents  in  every  OPERATION. 

dollar,  there  must  be  100  times  as  many  cents        67  X  100  —  6700 
as  dollars  in  the  given  sum.     But  to  multiply        A  US.  6700  cts. 
by  loo  we  annex  two  ciphers.     (Art.  57.) 

2.  In  $84,  how  many  mills? 

ANALYSIS. — For  a  like  reason,  there  are  84  X  1000  =  84000 
1000  times  as  many  mills  as  dollars,  or  10  Ans.  84000  mill*, 
times  as  many  mills  as  cents.  Hence,  the 

RULE. — To  reduce  dollars  to  cents,  multiply  them  by  100. 
To  reduce  dollars  to  mills,  multiply  them  by  1000. 
To  reduce  cents  to  mills,  multiply  them  by  10. 


198.  How  read  United  States  money  T 


UNITED     STATES     MONEY.  157 

NOTE. — Dollars  and  cents  are  reduced  to  cents ;  also  dollars,  cents, 
and  mills,  to  mills,  by  erasing  the  sign  of  dollars  ($,),  and  the  deci- 
mal point. 

2.  Reduce  $135  to  cents.  6.  Reduce  97  cents  to  mills. 

3.  Reduce  $368  to  mills.  7.  Reduce  $356.25  to  cents. 

4.  Reduce  $100  to  mills.  8.  Reduce  $780.375  to  mills. 

5.  Reduce  $1680  to  cents.  9.  Reduce  $800.60  to  mills. 

CASE    II. 

200.  To  reduce  Cents  and  Mills  to  Dollars, 
i.  In  6837  cents  how  many  dollars? 

ANALYSIS. — Since  100  cents  make  I  dollar,         OPERATION. 
6837  cents  will  make  as  many  dollars  as  100  is          I ) 68 137 
contained   times  in   6837,  and   6837-:- 100=68,  $68.37  -A.US. 

and  37  cents  over.    (Art.  79.) 

In  like  manner  any  number  of  mills  will  make  as  many  dollars  as 
looo  is  contained  times  in  that  number.  Hence,  the 

RULE. — To  reduce  cents  to  dollars,  divide  them  by  100. 

To  reduce  mills  to  dollars,  divide  them  by  1000. 

To  reduce  mills  to  cents,  divide  them  by  10.     (Art.  194.) 

NOTE. — The  first  two  figures  cut  off  on  the  right  are  cents,  the 
next  one  mills. 

2.  Reduce  1625  cts.  to  dols.  6.  Change  89567  cts.  to  dols. 

3.  Reduce  8126  m's  to  dols.  7.  Change  94283  m's  to  dols. 

4.  Reduce  10000  m's  to  dols.  8.  Change  85600  m's  to  cents. 

5.  Reduce  9265  m's  to  cents,  o.  Change  263475  m's  to  dols. 

10.  A  farmer  sold  763  apples,  at  a  cent  apiece :  ho\V 
many  dollars  did  they  come  to  ? 

u.  A  market  woman  sold  5  hundred  eggs,  at  2  cents 
each :  how  many  dollars  did  she  receive  for  them  ? 

12.  A  fruit  dealer  sold  675  watermelons  at  1000  mills 
apiece :  how  many  dollars  did  he  receive  for  them  ? 

199.  How  reduce  dollars  to  cents?  To  mills?  How  cents  to  mills  ?  Note.  How 
dollars  to  cents  and  mills?  aoo.  How  reduce  cents  and  mills  to  dollars?  JVofe 
What  are  the  fisares  cut  off? 


158  UtfllED     STATES     HONEY. 


ADDITION   OF   UNITED    STATES   MONEY. 

201.  United  Slates  Money,  we  have  seen,  is  founded 
upon  the  decimal  notation;  hence,  all  its  operations  ara 
precisely  the  same  as  the  corresponding  operations  in 
Decimal  Fractions. 

202.  To  find  the  Amount  of  two  or  more  Sums  of  Money. 

1.  What  is  the  sum  of  $45.625  ;  $109.07  ;  and  $450.137  ? 

ANALYSIS. —  Units  of    the  same  order  only  can  be  OPERATION. 

added  together.     For  convenience  in  adding,  we  there-  $45.625 

fore  write  dollars  under  dollars,  cents  under  cents,  etc.,  109,07 

with  the  decimal  points  in  a  perpendicular  line.     Begin-  450. 137 

Hing  at  the   ricrht,   we  add   the  columns  separately,  $604.8^2 
placing  the   decimal  point  in  the  amount   under  the 

points  in  the  numbers  added,  to  distinguish  the  dollars  from  cents 
and  mills.     (Art.  197.)    Hence,  the 

EULE. —  Write  dollars  under  dollars,  cents  under  cents, 
etc.,  and  proceed  as  in  Addition  of  Decimals. 

NOTE. — If  any  of  the  given  numbers  have  no  cents,  their  place 
should  be  supplied  by  ciphers. 

2.  A  man  paid  $13.62^  for  a  barrel  of  flour,  $25.25  for 
butter,  $9.75  for  coal :  what  did  he  pay  for  all  ? 

3.  A  farmer  sold  a  span  of  horses  for  $457.50,  a  yoke  of 
oxen  for  $235,  and  a  cow  for  $87.75  :  now  much  did  he 
receive  for  all  ? 

4.  What  is  the  sum  of  $97.87^;  $82.09;  $20. 12^? 

5.  What  is  the  sum  of  $81.06;  $69.18;  $67.16;  $7.13? 

6.  What  is  the  sum  of  $101.101;  $210.10^;  $450.27^? 

7.  Add  $7  and  3  cents;  $10;  6£  cents;  i8f  cents. 

8.  Add  $68  and  5  mills;  87^  cents;  31!  cents. 

9.  A  man  paid  $8520.75  for  his  farm,  $1860.45  for  hia 
stock,  $1650.45  for  his  house,  and  $1100.07  f°r  his  furni- 
ture :  what  was  the  cost  of  the  whole  ? 

•?nj.  What  is  pnld  of  operations  in  United  StiteR  Money?  203.  How  add 
Unite  J  States  Money?  Note.  If  any  of  the  numbers  have  no  cents,  how  proceed* 


UNITED     STATES     MONET.  159 

10.  A  lady  paid  $31^  for  a  dress,  $15  J-  for  trimmings, 
and  $7|-  for  making:  what  was  the  cost  of  her  dress  ? 

11.  A  grocer  sold  goods  to  one  customer  amounting  to 
$17.50,  to  another  $30.18^,  to  another  $21.06-},  and  to 
another  $51.73  :  what  amount  did  he  sell  to  all  ? 

12.  A  young  man  paid  $31.58  for  a  coat;  $11.63  f°r  a 
rest,  $14.11  for  pants,  $10.50  for  boots,  $7^  for  a  hat,  and 
$if  for  gloves :  what  did  his  suit  cost  him  ? 

SUBTRACTION   OF   UNITED   STATES   MONEY. 

203.  To  find  the  Difference  between  two  Sums  of  Money. 

1.  A  man  having  $1343.87^,  gave  $750.69  to  tlie  Patriot 
Orphan  Home :  how  much  had  he  left  ? 

ANALYSIS. — Since  the  same  orders  only  can  be  sub-  OPEBATIOX. 

traded  one  from  the  other,  for  convenience  we  write  $1343.875 

dollars  under  dollars,  cents  under  cents,  etc.     Begin-  750.69 

ning  at  the  right,  we  subtract  each  figure  separately,  &CQ.,  jgT 
and  place  the  decimal  point  in  the  remainder  under 
that  in  the  subtrahend,  for  the  same  reason  as  in  subtracting  deci- 
mals.   (Art.  190.)    Hence,  the 

KULE. —  Write  the  less  number  under  the  greater,  dollars 
under  dollars,  cents  under  cents,  etc.,  and  proceed  as  in 
Subtraction  of  Decimals.  (Art,  190.) 

NOTE. — If  only  one  of  the  given  numbers  has  cents,  their  place  in 
the  other  should  be  supplied  by  ciphers. 

2.  A  man  having  $861.73,  lost  $328.625  in  gambling: 
how  much  had  he  left  ? 

3.  If  a  man's  income  is$i75o,and  his  expenses  $1145.37!, 
how  much  does  he  lay  up  ? 

4.  A  gentleman  paid  $1500  for  his  horses,  and  $975 \  for 
his  carriage :  what  was  the  difference  in  the  cost  ? 

5.  A  merchant  paid  $2573^  for  a  quantity  of  tea,  und 
sold  it  for  $3158! :  how  much  did  he  make  ? 

203.  How  subtract  United  States  money?  Note.  If  either  number  has  no 
cents,  what  is  to  be  done  ? 


ICO  UNITED     STATES     MONEY. 

6.  If  I  pay  $5268  for  a  farm,  and  sell  it  for  84319.67, 
how  much  shall  I  lose  by  the  operation  ? 

7.  From  673  dols.  6J  cents,  take  501  dols.  and  10  cents. 

8.  From  ion  dols.  \z\  cents,  take  600  dols.  and  5  cents. 

9.  From  i  dollar  and  i  cent,  subtract  5  cents  5  mills. 
10.  From  i\  dols.  subtract  7^  cents. 

n.  From  500  dols.  subtract  5  dols.  5  cents  and  5  mills. 

12.  A  young  lady  bought  a  shawl  for  $35^,  a  dress  for 
$23^,  a  hat  for  $iof,  a  pair  of  gloves  for  $i£,  and  gave 
the  clerk  a  hundred  dollar  bill:  how  much  change  ought 
she  to  receive  ? 

MULTIPLICATION  OF  UNITED  STATES  MONEY. 

204.  To  multiply  United  States  Money. 

1.  What  will  9~£  yards  of  velvet  cost,  at  $i8-£  per  yard  ? 

ANALYSIS. — 9^  yards  will  cost  9^  times  as  much  as  $18.21; 

i   yard.    Now  9^  yds.=g.5  yds.,  and  $i8j=$i8.25.  _  - 
We  multiply  in  the  usual  way,  and  since  there  aro 
three,  decimal  figures  in  both  factors,  we  point  off  three 

in  the  product  for  the  same  reason  as  in  multiplying  I   425  ^ 

decimals.    (Art.  191.)    Hence,  the  $J73'375 

RULE. — Multiply,  and  point  off  the  product,  as  in  Multi- 
plication of  Decimals.  (Art.  191.) 

NOTES. — i.  In  United  States  Money,  as  in  simple  numbers,  the 
multiplier  must  be  considered  an  abstract  number. 

2.  If  either  of  the  given  factors  contains  a  common  fraction,  it  ia 
generally  more  convenient  to  change  it  to  a  decimal. 

2.  "What  will  65  barrels  of  flour  cost,  at  $11.50  a  barrel? 

3.  "What  will  145.3  pounds  of  wool  cost,  at  $i. 08  a  pound? 

4.  What  cost  75  pair  of  skates,  at  $3.87^  a  pair  ? 

5.  What  cost  63   gallons  of  petroleum,  at  95 £  centa 
a  gallon  ? 

6.  At  $4.17^  a  barrel,  what  will  no  barrels  of  apples  cost? 

804.  How  multiply  United  States  money  ?    Note,  What  must  the  multiplier  be  T 


UNITED     STATES     MONEY.  J.C1 

Perform  the  following  multiplications : 

7.  $5i$xi.9$.  ii.  $765.40$  x  6.05. 

8.  $6.07^- x  2.3 J.  12.  6.07  x. 008. 

9.  $10.05  x  6-$>  I3-  $-°°5  x  1000. 
10.  $100.031  x  3.105.                  14.  $1.01  ix. ooi. 

15.  What  cost  35  pounds  of  raisins,  at  i8f  cents  a 
pound  ? 

16.  What  cost  51  pounds  of  tea,  at  $1.15^  a  pound? 

17.  What  cost  150  gallons  of  milk,  at  37  £  cts.  a  gallon  ? 

18.  What  cost  12  dozen  penknives,  at  31^  cents  apiece? 

19.  What  cost  13  boxes  of  hutter,  each  containing  i6£ 
pounds,  at  37 £  cents  a  pound? 

20.  A  merchant  sold  12  pieces  of  cloth,  each  containing 
35  yards,  at  $4-|  a  yard :  what  did  it  come  to  ? 

21.  What  is  the  value  of  21  bags  of  coffee,  each  weigh- 
ing 55  pounds,  at  47^  cents  a  pound  ? 

22.  What  cost  55  boxes  of  lemons,  at  $3^  a  box  ? 

23.  The  proprietor  of  a  livery  stable  took  37  horses  to 
board,  at  $32  a  month:  how  much  did  he  receive  in  12 
months  ? 

DIVISION   OF   UNITED    STATES    MONEY. 

205.  Division  of  United  States  money,  like  simple  divi- 
sion, embraces  two  classes  of  problems : 

First.  Those  in  which  both  the  divisor  and  dividend  are  money. 

Second.  Those  in  which  the  dividend  is  money  and  the  divisor  is 
an  abstract  number,  or  regarded  as  such. 

In  the  former,  a  given  sum  is  to  be  divided  into  parts,  the  value  of 
each  part  being  equal  to  the  divisor ;  and  the  object  is  to  ascertain 
the  number  of  parts.  Hence,  the  quotient  is  times  or  an  abstract 
number.  (Art.  63,  «.) 

In  the  latter,  a  given  sum  is  to  be  divided  into  a  given  number  of 
equal  parts  indicated  by  the  divisor;  and  the  object  is  to  ascertain 
the  value  or  number  of  dollars  in  each  part.  Hence,  the  quotient  is 
money,  the  same  as  the  dividend.  (Art.  63,  6.) 

905.  What  docs  Division  of  U.  8.  money  embrace  T  The  flrst  class  ?  The  second  I 


162  UNITED     STATES     MONET. 

206.  To  Divide  Money  by  Money,  OP  by  an  Abstract  Number 

i.  How  many  barrels  of  flour,  at  $9.56  a  barrel,  can  be 
bought  for  $262.90  ? 

ANALYSIS.  —  At  $9.56  a  barrel,  $262.90  will  OPERATION. 

buy  as  many  barrels  as  $9.56  are  contained     $ 
times  in  $262.90,  or  27.5  barrels.     As  the  1912 

divisor  and  dividend    both    contain  cents, 


they  are  the  same  denomination  ;  therefore,  6602 

the  quotient  27,  is  a  whole  number.     Annex- 


ing a  cipher  to  the  remainder,  the  next  quo- 
tient figure  is  tenths.  (Art.  193,  Hem.) 
Hence,  the 

EULE.  —  Divide,  and  point  off  the  quotient,  as  in  Division 
of  Decimals.  (Arts.  64,  193.) 

NOTES.  —  i.  In  business  matters  it  is  rarely  necessary  to  carry  the 
quotient  beyond  mills. 

2.  If  there  is  a  remainder  after  all  the  figures  of  the  dividend  have 
been  divided,  annex  ciphers  and  continue  the  division  as  far  as  de- 
sirable, considering  the  ciphers  annexed  as  decimals  of  the  dividend. 

2.  If  63  caps  cost  $123.75,  what  will  i  cap  cost? 

3.  If  165  lemons  cost  $8.25,  what  will  i  lemon  cost? 

4.  Paid  $852.50  for  310  sheep  :  what  was  that  apiece  ? 

5.  If  356  bridles  cost  $1040,  what  will  i  bridle  cost? 

Find  the  results  of  the  following  divisions: 

6.  $i7.5o-r-$.i75  10.  $.oo5-r-$.o5 

7.  $365.07-^-  $1.01  n.  $ioi-=-$i.oi 

8.  $1000-1-25  cts.  12.  $500^  $.05 

9.  $.25-r$25  13.    $I200-H$.002 

14.  If  410  chairs  cost  $1216,  what  will  i  chair  cost? 

15.  At  $9$  a  ton,  how  much  coal  will  $8560  buy? 

16.  When  potash  is  $120.35  Per  ton,  how  much  can  be 
bought  for  $35267.28  ? 

17.  If  2516  oranges  cost  $157.25,  what  will  one  cost? 
*8.  Paid  $273.58  for  5000  acres  of  land:  what  was  that 

per  acre  ? 

ao6.  flow  divide  money?    Note.  If  there  ie  a  remainder,  bow  proceed  ? 


UNITED     STATES     MONET. 


COUNTING-ROOM   EXERCISES. 


163 


207.  The  Ledger  is  the  principal  look  of  accounts 
kept  by  business  men.    It  contains  a  brief  record  of  their 
monetary  transactions.     All  the  items  of  the  Day  Book  are 
transferred  to  it  in  a  condensed  form,  for  reference  and 
preservation,  the  debits  (marked  Dr.)  being  placed  on  the 
left,  and  the  credits  (marked  Or.)  on  the  right  side. 

208.  Balancing  an  Account  is  finding  the  differ- 
ence between  the  debits  and  credits. 

Balance  the  following  Ledger  Accounts : 

('•) 


DR. 

CR. 

DR. 

CR. 

$8645.23 

$2347.19 

$76421.26 

$43261.47 

160.03 

141.07 

2406.71 

728.23 

2731.40 

2137.21 

724.05 

6243.41 

4242.25 

3401.70 

86025.21 

75-69 

324-31 

2217.49 

9307.60 

53268.75 

33i3-i7 

168.03 

685.17 

3102.84 

429.18 

329.17 

61.21 

456.61 

4536.20 

2334-67 

7824.28 

32921.70 

641.46 

4506.41 

60708.19 

6242.09 

182.78 

239.06 

1764.85 

20374-34 

2634.29 

2067.12 

32846.39 

290.25 

3727-34 

675.89 

385-72 

4536.68 

840.68 

1431.07 

23.64 

42937-74 

6219.77 

4804.31 

6072.77 

8i9-35 

4727.91 

6536.48 

50641.39 

30769.27 

23-45 

720.34 

2062.40 

8506.35 

650.80 

36-45 

301-53 

97030.48 

3267.03 

7050.63 

26.47 

876.20 

680.47 

904.38 

805.03 

89030.50 

64.38 

78.05 

2403.67 

384.06 

8350.60 

307.63 

10708.79 

70207.48 

207.  What  is  the  ledger  ?  What  does  it  contain  ?   How  balance  an  account  ? 


104  BILLS. 

MAKING    OUT     BILLS. 

209.  A  Hill  is  a  written  statement  of  goods  sold,  ser- 
vices rendered,  etc.,  and  should  always  include  the  price 
of  each  item,  the  date,  and  the  place  of  the  transaction. 

A  Sill  is  Htceipted,  when  the  person  to  whom  it 
is  due,  or  his  agent,  writes  on  it  the  words  "  Received 
payment,"  and  his  name. 

209,  «.  A  Statement  of  Account  is  a  copy  of  the 
items  of  its  debits  and  credits. 

NOTES. — i.  The  abbreviation  Dr.  denotes  debit  or  debtor;  Or., 
credit  or  creditor ;  per,  by  ;  the  character  @,  stands  for  at.  Thus,  5 
books,  @  3  shillings,  signifies,  5  books  the  price  of  which  is  3  shil- 
lings apiece. 

2.  The  learner  should  carefully  observe  the  form  of  Bills,  the 
place  where  the  date  and  the  names  of  the  buyer  and  seller  are 
placed,  the  arrangement  of  the  items,  etc. 

Copy  and  find  the  amount  due  on  the  following  Bills : 

(i.)  PHILADELPHIA,  March  ist,  1872. 

Hon.  HENRY  BARNARD, 

To  J.  B.  LIPPINCOTT  &  Co.,  Dr. 


For  6  Webster's  Dictionaries,  4to.,  @  $12.50 
"    8  reams  paper,                             @    $3.75 
"  36  slates,                                       @    $0.27  | 

CKEDIT. 

$75 
30 
9 

oo 
oo 

72 

$114 

72 

By  10  School  Architecture, 
"     8  Journals  of  Education, 

@    $545 
@    $3-75 

$54 
30 

50 

oo 

| 

50 

Balance,      \ 

i      $30       22 

209.  What  Is  a  bill  ?  209,  a.  A  statement  of  account  ?  Where  is  the  date 
placed?  (See  form.)  The  name  of  buyer?  Seller?  How  is  the  payment  of  a 
bill  shown  t  Note.  What  does  Dr.  denote  f  Cr.  t 


BILLS.  165 

NEW  YORK,  April  id,  1871. 


Mrs.  W.  C.  GAEFIELD, 


Bought  of  A.  T.  STEWART. 


28  yds.  silk, 
35  yda.  table  liiien, 
6  pair  gloves, 
43-^  yds.  muslin, 
I  dew.  pr.  cotton  hose, 

Amount, 

@  $3-50 

@$2.I2i 

©$i-75 
@  $0.33 
<&  $0.80 

! 

Received  Payment, 

A.  T.  STEWART. 

(3-)  KEW  ORLEANS,  J/ay  5^,  1872. 

GEORGE  PEABODY,  ESQ., 

Bought  of  JACOB  BARKER. 


Feb.     i 

I? 
March  3 

25 
April  30 

75  bis.  pork,                     @,  $25.00 
160  bis.  flour.                     @    $8.75 
500  gals,  molasses,            @    $0.93 
75  boxes  raisins,               @    $5.37  i 
256  gals,  kerosene,            @    §0.87  j 

Amount, 

| 

Received  Payment, 


J.  BARKER, 
By  JOHN  HOWARD. 


(A\  BOSTON,  May  23^,  1871. 

Messrs.  FAIRFIELD  &  WEBSTER, 

To  H.  W.  HALL,  Dr. 


For    319  yds.  broadcloth, 
"     416  yds.  cassimere, 
"    I  no  yds.  mnslin, 
"      265  yds.  ticking, 

Amount, 

@  $5-87i 

@  $2.10 
(^  $0.28 
®  $047 

Received  Payment  by  Note, 

GEORGE  ANDEEBON, 
for  H.  W.  HALL- 


166  BILLS. 

5.  James  Brewster  bought  of  Horace  Foote  &  Co.,  New 
Haven,  May  i6th,  1867,  the  following  items:  175  pounds 
of  sugar,  at   17  cents ;  5  gallons  of  molasses,  at  63  cents ; 
3  boxes  of  raisins,  at  $6£ ;  15  pounds  of  tea,  at  $i| :  what 
was  the  amount  of  his  bill? 

6.  Georgo  Bliss  &  Co.  bought  of  James  Henry,  Cincin- 
nati, June  3d,  1867,  1625  bushels  of  wheat,  at  $1.95 ;  130 
barrels  of  flour,  at  $n;  265  pounds  of  tobacco,  at  48 
cents;  and  1730  pounds  of  cotton,  at  2 7^  cents:  what  was 
the  amount  of  their  bill  ? 

7.  Pinkney  &  Brother  sold  to  Henry  Rutledge,  Rich- 
mond, July  isth,  1867,  i  shawl,  $450;  19  yards  of  silk,  at 
$3.63  ;  1 6  yards  point  lace,  at  Si  i ;  6  pair  gloves,  at  $2.05  ; 
and  12  pair  hose,  at  87-^  cents:  required  the  amount. 

8.  Bought    37    Greek    Readers,   at    $1.85;    60    Greek 
Grammars,  at  $1.45;  75  Latin  Grammars,  at  $1.38;  25 
Virgils,  at  $3.62  ;  14  Iliad,  at  $3.28:  required  the  amount 

BUSINESS   METHODS. 

210.  Calculations  in  United  States  money  are  the  same 
as  those  in  Decimals ;  consequently,  in  ordinary  business 
transactions  no  additional  rules  are  required.  By  Reflec- 
tion and  Analysis,  however,  the  operations  may  often  be 
greatly  abbreviated.  (Arts.  100,  211-214.) 

1.  What  will  265  hats  cost,  at  $7  apiece  ? 

ANALYSIS. — 265  hats  will  cost  265  times  as  much  as  one  hat ;  arid 
$7x265  =  11855. 

Or  thus :  At  $i  each,  265  hats  will  cost  $265  ;  hence,  at  $7,  they 
will  cost  7  times  $265,  and  $265  x  7=$i8ss. 

NOTES. — i.  Here  the  price  of  one  and  the  number  of  articles  are  given, 
to  find  their  cost;  hence,  it  is  a  question  in  Multiplication.  (Art.  51.) 

2.  A  mechanic  sold  12  ploughs  for  $114:  what  was  the 
price  of  each  ? 

ANALYSIS. — The  price  of  i  plough  is  A  as  much  as  that  of  13 
ploughs;  and  $114-^12  =  19.50. 


COUNTING-ROOM     EXERCISES.  167 

NOTE. — 2.  Here  the  number  of  articles  and  their  cost  are  given, 
to  find  the  price  of  one,  which  is  simply  a  question  in  Division. 

3.  A  farmer  paid  $921.25  for  a  number  of  sheep  valued 
at  $2.75  apiece:  how  many  did  he  buy? 

ANALYSIS. — At  $2.75  apiece,  $921.25  will  buy  as  many  sheep  as 
$2.75  is  contained  times  in  $921.25  ;  and  $g2i.25-r-$2.75  =  335. 

NOTE. — 3.  Here  the  whole  cost  and  the  price  of  one  are  given,  to 
find  the  number,  which  is  also  a  question  in  Division. 

4.  What  will  287  chairs  cost,  at  $2f  apiece? 

5.  What  will  75  sofas  cost,  at  $57.50  apiece? 

6.  What  will  350  table  books  cost,  at  12-^  cents  apiece? 

7.  What  cost  119  tons  of  coal,  at  87^  a  ton  ? 

8.  Paid  $84  for  252  pounds  of  butter:  what  was  that  a 
pound  ? 

9.  If  8£  cords  of  wood  cost  $46.06,  what  will  i  cord  cost? 

10.  If  46  acres  of  laud  are  worth  $1449,  what  is  i  acre 
worth  ? 

11.  How  many  Bibles  at  $5^  can  be  bought  for  $561  ? 

12.  How  many  vests  at  $8^,  can  be  bought  for  $1262.25  ? 

13.  How  much  will  it  cost  a  man  a  year  for  cigars, 
allowing  he  smokes  5  a  day,  averaging  6^  cents  each,  and 
365  days  to  a  year? 

14.  A  lady  bought  18  yards  of  silk,  at  $2.12^  a  yard;  14 
yards  of  delaine  at  63  cents;  12  skeins  of  silk,  at  6\  cents 
a  skein :  what  was  the  amount  of  her  bill  ? 

15.  What  will  it  cost  to  build  a  railroad  265  miles  long, 
at  $11350  per  mile? 

1.6.  A  farmer  sold  his  butter  at  34  cents  a  pound,  and 
received  for  it  $321.67^:  how  many  pounds  did  he  sell  ? 

17.  The  cheese  made  of  the  milk  of  53  cows  in  a  season 
was  sold  for  $1579.40,  at  20  cents  a  pound:  how  many 
pounds  were  sold ;  and  what  the  average  per  cow  ? 

1 8.  A  merchant  sold  3  pieces  of  muslin,  each  containing 
45  yards,  at  40^  cents  a  yard,  and  took  his  pay  in  wheat 
at  $i    a  bushel:  how  much  wheat  did  he  receive? 


168  COUNTING-ROOM     EXEKCISES. 

19.  If  a  man  earns  $9^  a  week,  and  spends  $4^,  ho\« 
much  will  he  lay  up  in  52  weeks  ? 

20.  If  a  man  drinks  3  glasses  of  liquor  a  day,  costing  10 
cents  a  glass,  and  smokes  5  cigars  at  6  cents  each,  how 
many  acres  of  land,  at  $i£  an  acre,  could  he  buy  for  the 
sum  he  pays  for  liquor  and  cigars  in  35  years,  allowing 
365  days  to  a  year  ? 

21.  A  grocer  bought  175  boxes  of  oranges,  at  $6.37^, 
and  sold  the  lot  for  $637.50:  what  did  he  make,  or  lose? 

211.  To  find  the  Cost  of  a  number  of  articles,  the  Price  of 
one  being  an  Aliquot  part  of  $f. 

i.  What  will  168  melons  cost,  at  12^  cents  apiece  ? 

ANALYSIS. — By  inspection  tlio  learner  will  perceive 
that  12^  cents=|  dollar.     Now.  at  i  dollar  apiece,        OPERATION. 
168  melons  will  cost  $168.     Bat  the  price  is  i  of  a      8)£6S 
dollar  apiece;   therefore,  they  will  cost  \  of  $168,          $21  Ans. 
which  is  $21      Hence,  the 

RULE. —  Take  such  a  part  of  the  given  number  as  indi- 
cated by  the  aliquot  part  of  &  i,  expressing  the  price  <f  one; 
the  result  will  be  the  cost.  (Arts.  105,  270.) 

2.  "What  will  265  pounds  of  raisins  cost,  at  25  cents? 

3.  What  cost  195  pounds  of  butter,  at  33 £  cents? 

4.  What  cost  352  skeins  of  silk,  at  6£  cents  ? 

5.  What  cost  819  shad,  at  50  cents  apiece? 

6.  At  i2\  cents  a  dozen,  what  will  100  dozen  eggs  cost  ? 

7.  At  20  cents  a  yard,  what  will  750  yards  of  calico  cost  ? 

8.  At  i6|  cents,  what  will  1250  pine  apples  come  to? 

9.  What  cost  1745  yards  of  delaine,  at  33 \  cents  ? 
10.  At  8£  cents  apiece,  what  will  375  pencils  cost? 
u.  At  25  cents  apiece,  what  will  1167  slates  cost? 

12.  How  much  will  175  dozen  eggs  cost,  at  20  cts.  a  doz.  ? 
13    What  will  219  slates  cost,  at  12$  cents  each? 

14.  At  i6|  cts.  apiece,  what  will  645  melons  cost? 

15.  At  33!  cts.  apiece,  how  much  will  347  penknives  cost  ? 


COUNTING-BOOM     EXERCISES.  169 

212.  To  find  the  Cost  of  a  number  of  articles,  the  Price  of 

one  being  $1  plus  an  Aliquot  part  of  $1. 

16.  What  cost  275  pounds  of  tea,  at  $1.25  a  pound? 

ANALYSIS. — The  price  of  i  pound  $1.25=$!  +  $i.    At       4)275 
$i  a  pound,  the  cost  of  275  pounds  would  be  $275.    But          AflTT 

the  price  is  fi  +  $.V;  therefore  275   pounds  will  cost  ;— 

once  $275  +  i  of  $275  =  $343-75-    Hence,  the  $343-75 

RULE. — To  the  number  of  articles,  add  \,  -J,  £  of  itself, 
as  the  case  may  be;  the  sum  will  be  their  cost. 

17.  At  $1.25  per  acre,  what  will  168  acres  of  land  cost? 

18.  At  $1.50  apiece,  what  will  365  chairs  cost  ? 

19.  What  cost  512  caps,  at  $i£  apiece  ? 

20.  At  $i.i6|  apiece,  what  cost  12  dozen  fans? 

21.  A  man  sold  200  overcoats  at  a  profit  of  $1.20  apices: 
how  much  did  he  make  ? 

213.  To  find  the  Number  of  articles,  the  Cost  being  given, 

and  the  Price  of  one  an  Aliquot  part  of  $1. 

22.  How  many  spellers,  at   \z\  cents  apiece,  can  be 
bought  for  $25  ? 

ANALYSIS. — 12^  cents=$^;  therefore  $25  will     $. I2i— :<pj 
buy  as  many  spellers  as  there  are  eighths  in  $25,     $25 -:-$-£  =  200 
and  25-^-^=200.     Ans.  zoo  spellers.     Hence,  the   ' 

RULE. — Divide  the  cost  of  the  whole  by  the  aliquot  part 
0/$i,  expressing  the  price  of  one. 

23.  How  many  yards  of  flannel,  at  50  cents,  can  be 
bought  for  $6.83  ? 

24.  How  many  pounds  of  candy,  at  33^  cents,  can  be 
purchased  with  $375  ? 

25.  Paid  $450  for  cocoa-nuts  which  were  25  cents  each, 
how  many  were  bought  ? 

26.  At  20  ,?ents  each,  how  many  pine-apples  can  be  pur- 
chased for  $538  ? 

211.  How  fina  the  cos."  of  articles  by  aliquot  parts  of  $i  ?  212.  How  when  the 
price  is  $i  plus  an  aliquot  part  of  $i»  213.  How  flnd  the  number  of  article^. 
when  cost  i#  given,  nnd  the  nrice  of  one  i?  an  r.liquot  part  of  $i  ? 


170  COUtfTING-ROOil     EXEKC1SES. 

214.  To  find  the  Cost  of  articles,  sold  by  the  100,  or  1000. 

1.  What  will  1765  oranges  cost,  at  $6.125  a  hundred? 

ANAIYBIS. — At    $6.125    f°r    each    orange,    1765         OPERATION. 
oranges  would  cost  1765  times  $6  125,  and  $6  125  x  $6.125 

1765  =  110810.625.     But   the   price   is   $6.125  tor  a  1765 

hundred  ;   therefore  this  product  is  100  times  too       $108  1062? 
large.     To  correct  it,  we  divide  by  100,  or  remove 
the  decimal  point  two  places  to  the  left.     (Art.  181.) 

In  like  manner  if  the  price  is  given  by  the  1000,  we  multiply  the 
price  and  number  of  articles  together,  and  remove  the  decimal  point 
in  the  product  three  places  to  the  left,  which  divides  it  by  1000. 
Heuce,  the 

RULE.  —  Multiply  the  price  and  number  of  articles 
together,  and  divide  the  product  by  100  or  1000,  as  the 
case  may  require.  (Art.  181.) 

NOTE  —In  business  transactions,  the  letter  C  is  sometimes  puv 
for  hundred;  and  M  for  thousand. 

2.  What  will  4532  bricks  cost,  at  $17.25  per  M.  ? 

3.  What  cost  1925  pounds  of  maple  sugar,  at  $12.50  per 
hundred? 

4.  What  cost  25268  feet  of  boards,  at  $31.25  per  thous- 
and ? 

5.  At  $5f  per  hundred,  how  much  will  20345  pounds 
of  flour  come  to  ? 

6.  At  $6.25  per  hundred,  what  will  19263  pounds  of 
codnsh  come  to  ? 

7.  What  cost  10250  envelopes,  at  $3.95  per  thousand? 

8.  What  cost  1275  oysters,  at  $1.75  per  hundred? 

9.  What  cost  13456  shingles,  at  $7.45  per  M.  ? 
io.  What  cost  82  rails,  at  $5!  a  hundred  ? 

n.  What  cost  93  pine  apples,  at  $15.25  a  hundred? 
12.  What  cost  355  feet  of  lumber,  at  $45  per  thousand  9 


214.  How  find  the  cost  of  articles  eold  by  the  100  or  1000  f    Note.  What  does  C 
etaadfor?     ThatM? 


COMPOUND    NUMBERS. 

215.  Simple  Numbers  are  those  which  contain 
units  of  one  denomination  only ;  as,  three,  five,  2  oranges, 
4  feet,  etc.     (Art.  101.) 

216.  Compound  Numbers  are  those  which  con- 
tain  units  of  two  or  more  denominations  of  the  same 
nature;  as,  5  pounds  and  8  ounces;  3  yards,  2  feet  and  4 
inches,  etc. 

But  the  expression  2  feet  and  4  pounds,  is  not  a  com- 
pound number;  for,  its  units  are  of  unlike  nature. 

NOTES. — i.  Compound  Numbers  are  restricted  to  the  divisions  of 
Money,  Weights,  and  Measures,  and  are  often  called  Denominate 
Numbers. 

2.  For  convenience  of  reference,  the  Compound  TaUes  are  placed 
together.  If  the  teacher  wishes  to  give  exercises  upon  them  as  they 
are  recited,  he  will  find  examples  arranged  in  groups  ooi  responding 
with  the  order  of  the  Tables  in  Arts.  276,  279. 


MOIS'EY. 

217.  Money  is  the  measure  or  standard  of  value.  It 
is  often  called  currency,  or  circulating  medium,  and  is  of 
two  kinds,  metallic  and  paper. 

Metallic  Money  consists  of  stamped  pieces  of  metal, 
called  coins.  It  is  also  called  specie,  or  specie  currency. 

Paper  Money  consists  of  notes  or  Mils  issued  by  the 
Government  and  Banks,  redeemable  in  coin.  It  is  often 
called  paper  currency. 

ais.  Wliat  are  simple  numbers  ?  216.  Compound?  Give  an  example  of  each. 
Note.  To  what  p.-e  compound  numbers  restricted?  217.  What  is  moacy?  Me- 
tallic money  ?  Paper  money  ? 


172  COMPOUND     NUMBERS. 


UNITED    STATES    MONEY. 

218.  United  States  Money  is  the  national  cur- 
rency of  the  United  States,  and  is  often  called  Federal 
Money.    Its  denominations  are  eagles,  dollars,  dimes,  cents, 
and  mills., 

TABLE. 

10  mills  (in)  are  i  cent,  •    ct. 

10  cents          "    i  dime, d. 

10  dimes         "    i  dollar,  ...        -  dol.  or  §. 

10  dollars       "    i  eagle  -        -        -        -        -    Z7. 

219.  The  Metallic  Currency  of  the  United  States 
consists  of  gold  and  silver  coins,  and  the  minor  coins.* 

1.  The  gold  coins  are  the  double  eagle,  eagle,  half  eagle,  quarter 
eagle,  three  dollar  piece,  and  dollar. 

The  dollar,  at  the  standard  weight,  is  the  unit  of  value. 

2.  The  silver  coins  are  the  dollar,  "trade"  dollar,  half  dollar, 
quarter  dollar,  and  dime. 

3.  The  minor  coins  are  the  $-cent  and  ycent  pieces,  and  the  cent. 

220.  The  iveiyht  and  jmrity  cf  the  coins  of  the 
United  States  are  regulated  by  the  laws  of  Congress. 

1.  The  standard  weight  of  the  gold  dollar  is  25.8  gr. ;  of  the  quar- 
ter eagle,  64  5  gr. ;  of  the  3-dollar  piece,  77.4  gr. ;  of  the  half  eagle, 
129  gr. ;  of  the  eagle,  258  gr. ;  of  the  double  eagle,  516  pr. 

2.  The  we'g7it  of  the  dollar  is  412  J  grains,  Troy;  the  "trade" 
dollar,  420  grains  ;  the  half  dollar,  i2i  grams  ;  the  quarter  dollar 
and  dime,  one-half  and  one-fifth  the  weight  of  the  half  dollar. 

3.  The  weight  of  the  5-cent  piece  is  77. 16  grains,  or  5  grams ;  of 
the  3-cent  piece,  30  grains ;  of  the  cent,  48  grains. 

4.  The  standard  purity  of  the  gold  and  silver  coins  is  nine-tenths 
pure  metal,  and  one-tenth  alloy.    The  alloy  of  gold  coins  is  silver  &nd 
copper;  the  silver,  by  law,  is  not  to  exceed  one-tenth  of  the  whole 
alloy.    The  alloy  of  silver  coins  is  pure  copper. 

219.  Of  what  does  the  metallic  currency  of  tha  United  States  consist  T  What 
are  the  gold  coins  ?  The  silver  ?  The  minor  coins  1  220.  IIow  is  the  weight  and 
purity  of  United  States  coins  rejrulated  ? 

*  Act  ef  Congrats,  March  $d,  1^73. 


COMPOUND     NUMBERS.  173 

5.  The  five-cent  and  three-cent  pieces  are  composed  of  one-fourth 
nickel  anJ  tJiree-fourths  copper ;  the  cent,  of  95  parts  copper  and  5 
parts  of  tin  and  zinc.  They  are  known  as  nickel,  and  bronze  coins. 

NOTES. — i.  The  Trade  dollar  is  so  called,  because  of  its  intended 
use  for  commercial  purposes  among  the  great  Eastern  nations. 

2 .  The  gold  coins  are  a  legal  tender  in  all  payments;  the  silver  coins, 
for  any  amount  not  exceeding  $5  in  any  one  payment;  the  minor 
coins,  for  any  amount  not  exceeding  25  cents  in  any  one  payment. 

3.  The  diameter  of  the  nickel  5-cent  piece  is  two  centimeters,  and 
its  weight  5  grams.    These  magnitudes  present  a  simple  relation  of 
the  Metric  weights  and  measures  to  our  own. 

4.  The  silver  5-cent  and  3-cent  pieces,  the  bronze  2-cent  piece, 
the  old  copper  cent  and  half -cent  are  no  longer  issued.    Mills  were 
never  coined. 

221.  The  Paper  Currency  of  the  United  States 
consists  of  Treasury-notes  issued  by  the  Government 
known  as  Greenbacks,  and  Bank-notes  issued  by  Banks. 

NOTE. — Treasury  notes  less  than  $i,'are  called  Fractional  Cur- 
rency. 

ENGLISH    MONEY. 

222.  English  Money  is  the  national  currency  of 
Great  Britain,  and  is  often  called  Sterling  Money.  The 
denominations  are  pounds,  shillings,  pence,  and.  fat  things. 

TABLE. 

4  farthings  (qr.  or  far.)  are  i  penny,  -    - d. 

12  pence  "    i  shilling,    - 8. 

20  shillings  "    i  pound  or  sovereign,    -    -  £ 

21  shillings  "    i  guinea,      ------  g. 

NOTES. — i.  The  gold  coins  are  the  sovereign  and  half  sovereign. 
The  pound  sterling  was  never  coined.  It  is  a  bank  note,  and  is 
represented  by  the  sovereign.  Its  legal  value  as  fixed  by  Congress  is 
$4.8665.  This  is  its  intrinsic  value,  as  estimated  at  the  U.  S.  Mint. 

What  is  the  alloy  of  gold  coins  ?  Of  silver?  221.  Of  what  does  U.  S.  paper 
currency  consist?  222.  English  money?  The  denominations?  The  Table? 
Note.  Is  the  pound  a  coin  ?  IIow  represented  ?  What  is  the  value  of  a  pound  ? 


174  COMPOUND    NUMBERS. 

2.  The  silver  coins  aro  the  crown  (53.);  the  hall-crown  (2s.  6d.), 
the  florin  (as ) ;   the  shilling  (i2d.) ;   the  six-penny,  four-penny,  and 
three-penny  pieces. 

3.  The  copper  coins  are  the  penny,  half-penny,  and  farthing. 

4.  Farthings  are  commonly  expressed  as  fractions  of  a  penny,  as  7$d. 

5.  The  oblique  mark  (/)  sometimes  placed  between  shillings  and 
pence,  is  a  modification  of  the  long  /. 

CANADA    MONEY. 

223.  Canada  Money  is  the  legal  currency  of  the 
Dominion  of  Canada.      Its  denominations  are    dollars, 
cents,  and  mills,  which  have  the  same  value  as  the  cor- 
responding denominations  of  U.  S.  money.     Hence,  all  the 
operations  in  it  are  the  same  as  those  in  U.  S.  money. 

NOTE. — The  present  system  was  established  in  1858. 

FRENCH     MONEY. 

224.  Frencli  Money  is  the  national  currency  of 
France.     The  denominations  are  the  franc,  the  decline, 
and  centime. 

TABLE. 

10  centimes        -        -        -        -    are  i  decime. 
10  declines          -        -        -        -      "    i  franc. 

NOTES. — i.   The  system  is  founded  upon  the  decimal  notation  ; 
hence,  all  the  operations  in  it  are  the  same  as  those  in  U.  S.  money. 

2.  The  franc  is  the  unit;  decimes  are  tenths  of  a  franc,  and 
centimes  hundredths. 

3.  Centimes  by  contraction  are  commonly  called  cents. 

4.  Decimes,  like  our  dimes,  are  not  used  in  business  calculations; 
they  are  expressed  by  tens  of  centimes.     Thus,  5  decimes  are  ex- 
jressed  by  50  centimes;  63  francs,  5  deci roes,  and  4  centimes  are 
written,  63.54  francs. 

5.  The  legal  value  of  the  franc  in  estimating  duties,  is  19.3  cents, 
its  intrinsic  value  being  the  same. 

Shilling  ?  Florin  ?  Crown  ?  How  are  farthings  often  written  ?  223.  What  la 
Canada  money  ?  Iff  denominations  ?  Their  valne  ?  224.  French  money  ?  Its 
denominations  »  The  Table  ?  JVbfc.  T\is  unit  ?  The  value  of  a  franc  ? 


WEIGHTS. 

225.  Weight  is  a  measure  of  the  force  called  gravity, 
by  which  all  bodies  tend  toward  the  center  of  the  earth. 

226.  Net  Weight  is  the  weight  of  goods  without  the 
bag,  cask,  or  box  which  contains  them. 

Gross  Weight  is  the  weight  of  goods  with  the 
bag,  cask,  or  box  in  which  they  are  contained. 

The  weights  in  use  are  of  three  kinds,  viz  :  Troy,  Avoir- 
dupois, and  Apothecaries'  Weight 

TROY    WEIGHT. 

227.  Troy  Weight  is  employed  in  weighing  gold, 
silver,  and  jewels.    The  denominations  are  pounds,  ounces* 
pennyweights,  and  grains. 

TABLE. 

24  grains    (gr.)     are  i  pennyweight,    -         -    -    pwt. 

20  pennyweights    "    i  ounce, oz. 

12  ounces  "    i  pound, lb. 

NOTE. — The  unit  commonly  employed  in  weighing  diamonds, 
pearls,  and  other  jewels,  is  the  carat,  which  is  equal  to  4  grains. 

228.  The  Standard  Unit  of  weight  in  the  United 
States,  is  the  Troy  pound,  which  is  equal  to  22.794377  cubic 
inches  of  distilled  water,  at  its  maximum  density  (39.83° 
Fahrenheit),*  the  barometer  standing  at  30  inches.    It  is 
exactly  equal  to  the  Imperial  Troy  pound  of  England,  the 
former  being  copied  from  the  latter  by  Captain  Kater.f 

NOTE. — The  original  element  of  weight  is  a  grain  of  wheat  taken 
from  the  middle  of  the  ear  or  head.  Hence  the  name  grain  as  a 
unit  of  weight. 

225.  What  is  weight ?    22i.  Troy  weight ?    The  denomination*  ?    The  table? 

227.  The  standard  unit  of  weight  ?     Note.  The  original  element  of  weight  T 

228.  Avoirdupois  weight  ?    The  denominations  ?    The  Tahle  ? 

*  Hassler.  t  Professor  A.  D.  Bache. 


176  COMPOUND    NUMBERS. 


AVOIRDUPOIS   WEIGHT. 

229.  Avoirdupois   Weight  is  used  in   weighing 
all  coarse  articles ;   as,  hay,  cotton,  meat,  groceries,  etc., 
und  all  metals,  except  gold  and  silver.    The  denominations 
are  tons,  hundreds, pounds,  and  ounces. 

TABLE. 

16  ouncss  (oz.}     •    -   are  i  pound, Ib. 

100  pounds,  -    -    -    -     "     I  hundredweight,     -    -    -  coot. 

20  cwt.,  or  2000  Ibs.,     "     i  ton, T. 

Tlie  following  denominations  are  sometimes  used : 

1000  ounces  are  i  cubic  foot  of  water. 
100  pounds   "     i  quintal  of  dry  fish. 
196  pounds   "     i  barrel  of  flour. 
200  pounds  "     i  barrel  of  fish,  beef,  or  pork, 
280  pounds   "     i  barrel  of  salt. 

NOTES. — i.  The  ounce  is  often  divided  into  halves,  quarters,  etc. 
2.  In  business  transactions,  the  dram,  the  quarter  of  25  Ibs.,  and 
t}\BJi7-kin  of  56  Ibs.,  are  not  used  as  units  of  Avoirdupois  Weight. 

HEM. — In  calculating  duties,  the  law  allows  112  pounds  to  a 
hundredweight,  and  custom  allows  the  same  in  weighing  a  few 
coarse  articles ;  as,  coal  at  the  mines,  chalk  in  ballast,  etc.  In  all 
departments  of  trade,  however,  both  custom  and  the  law  of  most  of 
the  States,  call  100  pounds  a  hundredweight. 

230.  The  Standard  Avoirdupois  pound  is  equal 
to  7000  grains  Troy,  or  the  weight  of  27.7015  cubic  inches 
of  distilled  water,  at  its  maximum  density  (39.83°  Fah.),  the 
barometer  being  at  30  inches.    It  is  equal  to  the  Imperial 
Avoirdupois  pound  of  England. 

231.  Comparison  of  Avoirdupois  and  Troy  Weight. 

7000  grains  equal  I  Ib.  Avoirdupois. 
5760       "  "       i  Ib.  Troy. 

437  $•      "          "      i  oz.  Avoirdupois. 
480         "          "      i  oz.  Troy. 


229.  To  what  Is  the  Avoirdupois  pound  equal?    230.  What  is  net  weight? 
Gross  weight? 


COMPOUND     NUMBERS.  177 


APOTHECARIES'    WEIGHT. 

232.  Apothecaries'  Weight  is  used  by  physicians 
in  prescribing,  and  apothecaries  in  mixing,  dry  medicines. 

20  grains  (gr.)  are  i  scruple,    -    -    -    -  sc.,  or  3 . 

3  scruples         "    i  dram, dr.,  or  3  . 

8  drams  "    I  ounce,      ....  oz.,  or  | . 

12  ounces          "    i  pound,      -    -    -    -  lb.,  or  ft>. 

NOTE. — The    only  difference    between    Troy  and  Apothecaries' 

weight  is  in  the  subdicision  of  the  ounce.    The  pound,  ounce,  and 
grain  are  the  same  in  each. 

232,  a.  Apothecaries9  Fluid  Mcasu re  is  use'd 
i;i  mixing  liquid  medicines. 

60  minims,  or  drops  (ttj.  or  gtt.)  are  i  fluid  drachm,    -    -  /3  . 

8  fluid  drachms  "    i  fluid  ounce,       -    -  /§  . 

16  fluid  ounces  "    i  pint,      -    .    .    .    .    0. 

8  pints  "    i  gallon,  -----    Cong. 

NOTE. — Gtt.  for  guttae,  Latin,  signifying  drops  ;    0,  for  oct'triits, 
Lathi  for  one-eighth ;  and  Cong,  conyiarium,  Latin  for  gallon. 


MEASURES   OF  EXTEISTSIOK 

233.  Extension  is  that  which  has  one  or  more  of 
the  dimensions,  length,  breadth,  or  thickness;   as,  line-*. 
Kiirfaces,  and  solids. 

A  line  is  that  which  has  length  without  breadth. 

A  surface  is  that  which  has  length  and  breadth  with- 
out thicJcness. 

A  solid  is  that  which  has  length,  breadth,  and  thickness. 

NOTE. — A  measure  is  a  conventional  standard  or  unit  by  which 
values,  weights,  lines,  surfaces,  solids,  etc.,  are  computed. 

234.  The  Standard  Unit  of  length  is  the  yard,  which  is  de- 
termined from  the  scale  of  Troughton,    at  the  temperature  of  62° 
Fahrenheit.     It  is  equal  to  the  British  Imperial  yard. 

232.  In  what  ia  Apothecaries'  Weiprht  used  ?  Table  ?  233.  What  is  extension  T 
A  line  ?  A  surface  ?  A  solid  ?  Not*.  A  measure  ?  214.  What  is  the  standard 
tmit  of  length  ? 


178  COMPOUND    NUMBERS. 

LINEAR   MEASURE. 

235.  Linear  Measure  is  used  in  measuring  that 
which  has  length  without  breadth;  as,  lines,  distances. 
It  is  often  called  Long  Measure.  The  denominations  are 
kagues,  miles,  furlongs,  rods,  yards,  feet,  and  inches. 

TABLE. 

12  inches  (in.)      are  i  foot,     ........  ft. 

3  feet  "    i  yard,    ........  yd. 

5  1  yds.  or  16.}  ft.  "    i  rod,  perch,  or  pole,    -    -    -  r.  or  p. 
40  rods  .         "    i  furlong,    .......  fur. 

8  fur.  or  320  rods  "    i  mile,    ......    -    -  m. 

3  miles  "    i  league,     .......  I. 

The  following  denominations  are  used  in  certain  cases  : 

4  in.  —  i  hand,  for  measuring  the  height  of  horses. 
9  in.  =  i  span. 

1  8  in.  =  i  cubit. 

6  ft.  =  i  fathom,  for  measuring  depths  at  sea. 
3.3  ft.  =  i  pace,*  for  measuring  approximate  distances. 

5  pa.  =  i  rod,  "  ^    " 
ill-statute  mi.  =  i  geographic  or  nautical  mile. 
60  geographic,  or 


,  ,, 

69i-itatute  m.  nearly,  }  =  I  deSree  on  the  e1uator- 
360  degrees  =  i  circumference  of  the  earth. 

A  knot,  used  in  measuring  distances  at  sea,  is  equivalent  to  a  nau- 
tical mile. 

(For  the  English  and  French  methods  of  determining  the  standard 
of  length,  see  Higher  Arith.) 

NOTES.  —  i.  The  original  element  of  linear  measure  is  a  grain  or 
kernel  of  barley.  Thus,  3  barley-corns  were  called  an  inch.  But 
the  barley-corn,  as  a  measure  of  length,  has  fallen  into  disuse. 

2.  The  inch  is  commonly  divided  into  halves,  fourths,  eighths,  or 
tenths  ;  sometimes  into  twelfths,  called  lines. 

3.  The  mile  of  the  Table  is  the  common  land  mile,  and  contains 
5280  ft.     It  i0  called  the  statute  mile,  because  it  is  recognized  by  law, 
both  in  the  United  States  and  England. 

235.  For  what  is  linear  measure  used  ?    The  denominations  ?    The  Table  T 
Note.  Tbe  original  element  of  linear  measure?  How  is  the  inch  commonly  divided? 
*  A  military  pace  or  step  is  variously  estimated  2\  and  3  foot. 


COMPOUND     NUMBERS.  179 

238.  The  Linear  Unit  employed  by  surveyors  is 
Gunter's  Chain,  which  is  4  rods  or  66  ft.  long,  and  is  sub- 

iivided  as  follows : 

•»• 

7.92  inches  (in.)  are  i  link, I. 

25  links  •'    i  rod  or  polo   -    -    -  r, 

4  rods  "    i  chain, ch. 

80  chains  "    i  milo m. 

NOTE. — Gfunter's  chain  is  so  called  from  the  name  of  its  inventor. 
Engineers  of  the  present  day  commonly  use  a  chain,  or  measuring 
tupj  100  feet  long,  each  foot  being  divided  into  tenths. 

CLOTH     MEASURE. 

237.  Cloth  Measure  is  used  in  measuring  those 
articles  of  commerce  whose  length  only  is  considered ;  as, 
cloths,  laces,  ribbons,  etc.     Its  principal  unit  is  the  linear 
yard.    This  is  divided  into  halves,  quarters,  eighths,  and 
sixteenths. 

TABLE. 

,  3  ft.  or  36  in.  are  i  yard,        -     -  -  -  yd. 

18  in.,  "  i  half  yard      -  -  -  i  yd. 

9  in.,  "  i  quarter  yard  -  -  i  yd. 

4'  in.,  "  i  eighth      "  -  -  i  yd. 

2\  in.,  "  i  sixteenth"  -  -  -^  yd. 

NOTES. — r.  The  old  Ells  Flemish,  English,  and  French,  are  no 
longer  used  in  the  United  States ;  and  the  nail  (2\  inches),  as  a  unit 
ot  measure,  is  practically  obsolete. 

2.  In  calculating  duties  at  the  Custom  Houses,  the  yard  is  divided 
into  tenths  and  hundredths. 

SQUARE    MEASURE. 

238.  Square  Measure  is  used  in  measuring  sur- 
faces, or  that  which  has  length  and  breadth  without  thick- 
ness; as,  land,   flooring,  etc.     Hence,  it  is  often  called 
land  or  surface  measure.    The  denominations  are  acres 
sjuare  rods,  square  yards,  square  feet,  and  square  inches. 

2-jS.  WTint  is  the  linear  unit  commonly  employed  by  surveyors  ?  z-^.  Cloth 
.Tisasnre?  Its  principal  unit?  How  is  the  vard  divided?  The  Table?  238.  Square 
measure?  The  denominations?  ThcTablo? 


180 


COMPOUND 


TABLE. 

144  sqnare  inches  (sq.  in.)  are  i  square  foot,    -    -  aj.  ft. 
9  square  feet  "    i  square  yard,   -    -  sq.  yd. 

3o^fcq.  yards,  or )  „    I   i  sq.  rod,   perch 

272!  sq.  feet,         j"  /or  pole,    -    -     -  sq.  r. 

160  square  rods  "    i  acre, A. 

640  acres  "    i  square  mile,  -    -  sq.  m. 

229.  The  Unit  of  Land  Measure  is  the  Acre. 
and  is  subdivided  as  follows : 


i  acre, 


-  P- 

-  nq.  c. 

-  A. 


625  sq.  links  are  i  pole  or  sq.  rod,  - 

16  poles  "    i  square  chain.    - 

10  sq.  chains,  or )    „ 
160  sq.  rods  ) 

NOTES. — i.  The  Rood  of  40  sq.  rods  is  no  longer  used  as  a  unit  of 
measure, 

2.  A  Square,  in  Architecture,  is  100  square  feet. 

239.  a.  The  public  lands  of  the  United  States  are  divided  into 
Toicmships,  Sections,  and  Quarter-sections. 

A  Township  is  6  miles  square,  and  contains  36  sq.  miles. 

A  Section  is  i  mile  square,  and  contains  640  acres. 

A  Quarter-section  is  160  rods  square,  and  contains  160  acres. 

240.  A  Square  is  a  rectilinear  figure 
which  has  four  equal  sides,  and  four  riff  fit 
any  lea.    Thus, 

A  Square  IncJt  is  a  square,  each  side 
of  which  is  i  inch  in  length. 

A  Square  Yard  is  a  square,  each  side 
of  which  is  i  yard  in  length. 

This  measure  is  called  Square  Measure, 
because  its  measuring  unit  is  a  square. 

241.  A  Rectangle  or  Rectangular  Figure  is  one  which 
!'.as/<??/r  sides  and  four  right  angles.    When  all  the  sides  are  equal, 
it  is  called  a  square;  when  the  opposite  sides  only  are  equal.it  is 
tailed  an  oblong  or  parallelogram. 

242.  The  A  rea  of  a  figure  is  the  quart  tity  of  surface  it  contains, 
end  is  often  called  its  superficial  contents. 

Note.  How  are  the  Government  lands  divided?  How  ranch  land  in  n  town- 
chip  ?  In  a  section  ?  In  a  quarter-section  ?  240.  What  is  a  square  ?  A  square 
inch  ?  A  ponare  yard  ?  Why  is  sqnare  measure  f>o  called  ?  241.  What  is  a  rect- 
angular  flgpre?  242.  What  is  the  area  of  a  figure? 


9  eq.  ft.  =  i  sq.  yd. 

COMPOUND     NUMBERS.  181 

243.  The  n rca  of  all  rectangular  surfaces  is  found  by  multiplying 
Hie  length  and  breadth  together. 

244.  The  area  and  one  side  being  given,  the  otlier  side  is  found 
y  dividing  the  area  by  the  given  side.  (Art.  93.) 


CUBIC     MEASURE. 

245.  Cubic  Measure  is  used  in  measuring  solids, 
or  that  which  has  length,  breadth,  and  thickness;  us, 
timber,  boxes  of  goods,  the  capacity  of  rooms,  ships,  etc. 
Hence  it  is  often  called  Solid  Measure.  The  denomina- 
tions are  cords,  tons,  cubic  yards,  cubic  feet,  and  cubic  inches. 

TABLE. 

1728  cubic  inches  (cu.  in.)  are  i  cubic  foot,       cu.  ft. 
27  cubic  feet  "    I  cubic  yard,      cu.  yd. 

128  cubic  feet  "    i  cord  of  wood,  G'. 

245,  ft»  A  Cord  of  wood  is  a  pile  8  ft.  long,  4  ft.  wide,  and  4 
ft.  high;  for  8  X4  X4=i28. 

A  Cord  Foot,  is  one  foot  in  length  of  such  a  pile  ;  hence,  8  cord 
ft.  =  i  cord  of  wood. 

245,  b.  A  Register  Ton  is  the  standard  for  estimating  the 
capacity  or  tonnage  of  vessels,  and  is  100  cu.  ft. 

A  Shippiny  Ton,  used  in  estimating  cargoes,  in  the  U.  S.,  is 
<o  cu.  ft. ;  in  England,  42  cu.  ft. 

NOTE. — The  ton  of  40  ft.  of  round,  or  50  ft.  of  hewn  timber  is  sel. 
ilora  or  never  used. 

246.  A  Cube  is  a  regular  solid  bounded  3xs*3=27  cu.  ft 
by  six  equal  squares  called  its  faces.    Hence, 

its  length,  breadth,  and  thickness  are  equal 
to  each  other.  Thus, 

A  Cubic  Inch  is  a  cube,  each  side  of 
v.-hich  is  a  square  inch  ;  a  Cubic  Yard  is  a 
cube,  each  side  of  which  is  a  square  yard,  etc. 

This  Measure  is  called  Cubic  Measure, 
because  its  measuring  unit  is  a  cube. 


• 


243.  How  is  the  area  of  rectangular  surfaces  found  ?  245.  Cubic  measure  ?  The 
denominations  ?  Table  ?  Note.  Describe  a  cord  of  wood  ?  A  cord  foot  ?  A 
register  ton  ?  Shipping  ton  ?  t+6.  What  is  a  cube  ?  A  cubic  inch  ?  Yard  i 


182  COMPOUND     NUMBERS. 

247.  -4  rectangular  body  is  one  bounded  by  six  rectangular  sides, 
eacli  opposite  pair  beiug  equal  and  parallel;  as,  boxes  of  goods, 
blocks  of  hewn  stone,  etc. 

When  all  the  sides  are  equal,  it  is  called  a  cube  ;  when  the  opposite 
sides  only  are  equal,  it  is  called  a  parallclopiped. 

248.  The  contents  or  solidity  of  a  body  is  the  quantity 
of  matter  or  space  it  contains. 

249    The  contents  of  a  rectangular  solid  are  found,  by 
multiplying  the  length,  breadth,  and  thickness  together. 


MEASURES    OF    CAPACITY. 

250.  The  capacity  of  a  vessel  is  the  quantity  of  space 
included  within  its  limits. 

Measures  of  Capacity  are  divided  into  two  classes, 
dry  and  liquid  measures. 

DRY     MEASURE. 

251.  Dry  Measure  is  used  in  measuring  grain, 
fruit,  salt,  etc.    The  denominations  are  chaldrons,  bushels, 
pecks,  quarts,  said,  frints. 

TABLE. 

2  pints  (pt.)    -    -    are  i  quart, qt. 

8  quarts      -    .    -      "    I  peck, pk. 

4  pecks,  or  32  qts.,     "    I  bushel, bu. 

36  bushels  -    -    -  "i  chaldron,      ....  ch. 

252.  The    Standard   Unit  of  Dry  Measure  is  the 
bushel,  which  contains  2150.4  cubic  inches,  or  77.6274  Ibs. 
avoirdupois  of  distilled  water,  at  its  maximum  density. 

247.  What  is  a  rectangular  body?  When  all  the  sides  are  equal,  what  called? 
When  the  opposite  sides  only  are  equal  ?  248.  What  are  the  contents  of  a  solid 
body?  240.  How  find  the  contents  of  a  rectangular  solid?  zw.  The  rapacity 
of  a  vessel?  2*1.  Dry  measure?  The  denominations ?  The  Table?  252.  The 
standard  unit  of  dry  measure? 


COMPOUND     NUMBERS. 


183 


It  is  a  cylinder  i8£  in.  in  diameter,  and  8  in.  deep,  the 
same  as  the  old  Winchester  bushel  of  England.*  The 
British  Imperial  Bushel  contains  2218.192  cu.  inches. 

NOTES. — i.  The  dry  quart  is  equal  to  i\  liquid  quart  nearly. 
2.  The  chaldron  is  used  for  measuring  coke  and  bituminous  coal. 

253.  The  Standard  Eushel  of  different  kinds  of  grain,  seeds, 
etc.,  according  to  the  laws  of  New  York,  is  equal  to  the  following 
number  of  pounds : 


32  Ibs.  =  i  bu.  of  oats. 


44  Ibs.  =  i 
48  Ibs.  =  i 

55  Ibs.  r=  i 

56  Ibs.  =  i 


Timothy  seed. 
(  buckwheat,    or 
l      barley. 

flax-seed. 

rye. 


58  Ibs.  =  i  bu.  of  corn. 

( wheat,  peas, 
•<  potatoes,  or 
(  clover-seed. 
j  beans,  or  blue- 
I  grass  seed, 
loo  Ibs.  =  i  cental  of  grain. 


60  Ibs.  =  i 


62  Ibs.  =  i 


NOTES. — i.  The  cental  is  a  standard  recently  recommended  by 
the  Boards  of  Trade  in  New  York,  Cincinnati,  Chicago,  and  other 
large  cities,  for  estimating  grain,  seeds,  etc.  Were  this  standard  gen- 
erally adopted,  the  discrepancies  of  the  present  system  of  grain  deal- 
ing would  be  avoided. 

2.  BusJiels  are  cJuinged  to  centals,  by  multiplying  them  by  the 
number  of  pounds  in  one  bushel,  and  dividing  the  product  by  100. 
f  he  remainder  will  be  hundredth  of  a  cental. 


LIQUID     MEASURE. 

254.  Liquid  Measure  is  used  in  measuring  milk, 
wine,  vinegar,  molasses,  etc.,  and  is  often  called  Wine 
Measure.  The  denominations  are  hogsheads,  barrels,  gal- 
lons, quarts,  pints,  and  gills. 

TABLE. 

4  gills  (ffi )  are  i  pint, pt. 

2  pints  "  i  quart, qt. 

4  quarts         "  i  gallon,  ...    -     .    -  gal. 

31^  gallons       "  i  barrel, bar.  or  bbl. 

63  gallons        "  i  hogshead,       ....  Jihd. 

254.  Liquid  measure ?    The  denominations?    Table? 
*  Professor  A.  D.  Bache. 


1C4  COMPOUND     XUMBE11S. 

255.  The  Standard  Unit  of  Liquid  Measure  is  tha 
gallon,  which  contains    231   cubic   inches,  or   8.338   Ibs. 
avoirdupoia  of  distilled  water,  at  its  maximum  density. 
The  British  Imperial  Gallon  contains  277.274  cu.  inches. 

NOTES. — i.  The  barrel  and  Tiogshead,  as  units  of  measure,  are 
chiefly  used  in  estimating  the  contents  of  cisterns,  reservoirs,  etc. 

2.  Beer  Measure  is  practically  obsolete  in  this  country.  The  old 
beer  gallon  contained  282  cubic  inches. 

CIRCULAR    MEASURE. 

256.  Circular  Measure   is    used  in    measuring 
angles,  land,  latitude  and  longitude,  the  motion  of  the 
heavenly  bodies,  etc.     It  is  often  called  Angular  Measure. 
The  denominations  are  signs,  degrees,  minutes,  and  seconds. 

TABLE. 

60  seconds  (")  are  i  minute,        -  -  -  ' 

60  minutes         "    i  degree,         -  -  -  o  ,  or  deg. 

30  degrees          "    i  sign,     -        -  -  -  g. 

12  signs,  or  360°  "    i  circumference,  -  dr. 

NOTE. — Sij  is  are  used  in  Astronomy  as  a  measure  of  the  Zodiac. 

257.  A  Circle  is  a  plane  figure  bounded  by  a  curve  line,  every 
part  of  which  is  equally  distant  from  a  point   within  called  the 
center. 

The  Circumference  of  a  circle  is  the  curve  line  by  which  it  is 
bounded. 

The  Diameter  is  a  straight  line  dra%vn  through  the  center,  ter- 
minating at  each  end  in  the  circumference. 

The  IttKfius  is  a  straight  line  drawn 
from  the  center  to  the  circumference,  and 
is  equal  to  half  thu  diameter. 

An  Arc  is  any  part  of  the  circumfer- 
ence. 

In  the  adjacent  figure,  A  D  E  B  F  is 
the  circumferenca  ;  C  the  center ;  A  B  the 
diameter ;  C  A,  C  D.  C  E,  etc.,  are  radii ; 
A  D,  D  E,  etc.,  are  arcs. 

255.  The  standard  of  Liquid  measure  ?  Note.  What  of  Beer  measure  ?  256.  In 
\vhatisCircularmeasureused?  The  denominations?  The  Table?  257.  What 
It  a  circle?  The  circumference?  Diameter?  Radius?  An  arc? 


COMPOUND     XUMBEES.  1S£ 

253.   A  Plane  Angle  is  the  quantity  of  divergence  of  two 
etra-igat  lines  starting  from  the  same  point. 

Tne  Lines  which  form  the  angle  are 
called  the  sides,  and  the  point  from  which 
they  start,  the  vertex.  Thus,  A  is  the  ver- 
tex of  the  angle  B  A  C,  A  B  and  A  C  the 
sides. 

259.  A  Perpendicular  is  a  straight 
line  which  meets  another  straight  line  so  as 
to  make  the  two  adjacent  angles  equal  to 
each  other,  as  A  B  C,  A  B  D. 

Each  of  the  two  lines  thus  meeting  is 
perpendicular  to  the  other. 


C  ii  D 

260.  A  Ttiylit  Angle  is  one  of  the  two  equal  angles  formed 
by  the  meeting  of  two  straight  lines  which  are  perpendicular  to 
each  other.  All  other  angles  are  called  oblique. 

260.  (i.  The  Pleasure  of  an  angle  is  the  arc  of  a  circle  included 
between  its  two  sides,  as  the  arc  D  E,  in  Fig.,  Art.  258. 

261.  A  Dct/rec.  is  one  36oth  part  of  the  circumference  of  a  circle. 
It  is  divided  into  60  equa.1  parts,  called  minutes  j  the  minute  is 
divided  into  60  seconds,  etc.    Hence,  the  length  of  a  degree,  minute, 
etc.,  varies  according  to  the  magnitude  of  different  circles. 

The  length  of  a  degree  of  longitude  at  the  equator,  also  the  aver- 
age length  of  a  degree  of  latitude,  adopted  by  the  U.  S.  Coast  Sur- 
vey, is  69.16  statute  miles.     At  the  latitude  of  30°  it  is  59.81  miles, 
at  6o3  it  is  34.53  miles,  and  at  90°  it  is  nothing.* 

262.  A  Semi-circumference  is  half  a  circumference,  or  180°. 
A  Oiifidratit  is  one-fourth  of  a  circumference,  or  90°. 

263.  If  two  diameters  are  drawn  perpendicular  to  each  other, 
they  will  form  four  right  angles  at  the  center,  and  divide  the  cir- 
cumference into  four  equal  parts.     Hence, 

A  right  angle  contains  90°  ;  for  the  quadrant,  which  measures  it, 
is  an  arc  of  90°. 

238.  A  plane  angle  ?  The  sides  ?  The  vertex  ?  259.  A  perpendicular  ?  260.  A 
right  angle  ?  260,  a.  What  is  the  measure  of  an  angle  ?  261.  What  is  a  degree? 
Upon  what  does  the  length  of  a  degree  depend  ?  What  its  length  at  the  equator  ? 
262.  What  is  a  semi-circumference  ?  How  many  degrees  does  it  contain  ?  A 
quadrant?  263.  If  two  diameters  are  drawn  perpendicular  to  each  other,  what 
Is  the  result  ?  How  many  degrees  in  a  right  angle  ? 

*  Encyclopedia  Britannica, 


186  COMPOUND     X  UMBERS. 


MEASUREMENT    OF    TIME. 

264.  Time  is  a  portion  of  duration.  It  is  divided 
into  centuries,  years,  months,  weeks,  days,  hours,  minutes, 
and  seconds. 

TABLE. 


60  seconds  (sec.)                  are  i  minute 

-    m. 

60  minutes 

i  hour    - 

•    h. 

24  hours 

i  day 

-    d. 

7  days 

i  week  . 

-    w. 

365  days,  or        )                     „ 
52  w.  and  id.} 

i  common  year 

-    c.y. 

366  days                                  " 

i  leap  year    - 

-   i.y. 

12  calendar  months  (mo.)    " 

i  civil  year    - 

-  ?/• 

ico  years                                 " 

i  century 

-    c. 

. — In  most  business  transactions  30  days  are  considered  a 
Four  weeks  are  sometimes  called  a  lunar  month. 

265.  A  Civil  Year  is  the  year  adopted  by  govern- 
ment for  the  computation  of  time,  and  includes  both  com- 
mon and  leap  years  as  they  occur.  It  is  divided  into  12 
calendar  months,  as  follows : 


January  (Jan.)  ist  mo.,   31  d. 

February  (Feb.)  2d      "      28  d. 

March  (Mar.)  3d      "      31  d. 

April  (Apr.)  4th    "      30  d. 

May  (May)  5th    "      31  d. 

June  (Jane)  6th    "      30  d. 


July  (July)  7th  mo.,  31  d. 

August  (Aug.)  8th  "  31  d. 
September  (Sep.)  gth  "  30  d. 
October  (Oct.)  loth  "  31  d. 
November  (Nov.)  nth  "  30  d. 
December  (Dec.)  i2th  "  31  d. 


NOTES. — i.  The  following  couplet  will  aid  the  learner  in  remem- 
bering the  months  that  have  30  days  each : 

"  Thirty  days  hath  September, 
April,  June,  and  November." 

All  the  other  months  have  31  days,  except  February,  which  in 
common  years  has  28  days ;  in  leap  years,  29. 

266.  Time  is  naturally  divided  into  days  and  years. 
The  former  are  measured  by  the  revolution  of  the  earth  on 
its  axis ;  the  latter  by  its  revolution  around  the  sun. 

264.  What  is  time  T    The  denominations  f    The  Table  ? 


COMPOUND     LUMBERS. 


187 


267.  A  Solar  year  is  the  time  in  which  the  earth, 
starting  from  one  of  the  tropics  or  equinoctial  points,  re- 
volves around  the  sun,  and  returns  to  the  same  point.    It 
is  thenc3  called  the  tropical  or  equinoctial  year,  and  is 
equal  to  365d.  $h.  48m.  49.7  sec.* 

NOTE. — i.  The  excess  of  the  solar  above  the  common  year  is  6 
hours  or  \  of  a  day,  nearly  ;  hence,  in  4  years  it  amounts  to  about  I 
day.  To  provide  for  this  excess,  i  day  is  added  to  every  4th  year, 
which  is  called  Leap  year  or  Bissextile.  This  additional  day  is  given 
to  February,  because  it  is  the  shortest  month. 

2.  Leap  year  is  caused  by  the  excess  of  a  solar  above  a  common, 
year,  and  is  so  called  because  it  leaps  over  the  limit,  or  runs  on  I  day 
more  than  a  common  year. 

3.  Every  year  that  is  exactly  divisible  by  4,  except  centennial 
years,  is  a  ieap  year  ;  the  others  are  common  years.    Thus,  1868,  '72, 
etc.,  are  leap  years;  1869, '70, '71,  are  common.     Every  centennial 
year  exactly  divisible  by  400  is  a  leap  year ;  the  other  centennial 
j  ears  are  common.    Thus,  1600  and  2000  are  leap  years;  1700,  1800, 
and  1900  are  common. 

268.  An  Apparent  Solar  Dai/  is  the  time  between 
the  apparent  departure  of  the  sun  from  a  given  meridian 
and  his  return  to  it,  and  is  shown  by  sun  dials. 

A  Trite  or  Wean  solar  day  is  the  average  length  of 
apparent  solar  days,  and  is  divided  into  24  equal  parts, 
called  hours,  as  shown  by  a  perfect  clock. 

269.  A   Civil  Day  is  the  day  adopted  by  govern- 
ments for  business  purposes,  and  corresponds  with  the 
mean  solar  day.    In  most  countries  it  begins  and  ends  at 
midnight,  and  is  divided  into  two  parts  of  12  hours  each; 
the  former  being  designated  A.  M.;  the  latter,  p.  M. 

NOTE.  A  meridian  is  an  imaginary  circle  on  the  surface  of  the 
earth,  passing  through  the  poles,  perpendicular  to  the  equator. 
A.  M.  is  an  abbreviation  of  ante  meridies,  before  midday ;  P.  M.,  of 
post  meridies,  after  midday. 

266.  How  is  time  naturally  divided  ?    How  is  the  former  caused  ?   The  latter  ? 
*  Laplace^Somcrville,  BaSly's  Tables. 


188 


COMPOUND     NTJMBEIiS. 


MISCELLANEOUS    TABLES. 


12  things  are  i  dozen. 
12  doz.        "    i  gross. 

24  slieets  are  i  quire  of  paper. 
20  quires  "    i  ream. 

2  leaves  are  I  folio. 

4  leaves   "    I  quarto  or  4to. 

8  leaves   "    i  octavo  or  8vo. 


12  gr.?ss     are  i  great  gross. 
20  things    "    i  score. 

2  reams      are  i  bundle. 
5  bundles    "    i  bale. 

12  leaves  are  i  duodecimo  or  i2ma 
1 8  leaves   "    i  eighteen  rao. 
24  leaves   "    i  twenty-four  mo. 


NOTE. — The  terms  folio,  quarto,  octavo,  etc ,  denote  the  number 
of  leaves  into  which  a  sheet  of  paper  is  folded  in  making  books. 

270.  Aliquot  Parts  of  a  Dollar,  or  100  cents. 


50  cents  =  $i 
33!$  cents  =  $i 
25  cents  =  $i 
20  cents  =  $5 
16]  cents  =  $k 


12^  cents  = 

10    cents  — 

8\  cents  = 

64  cents  = 

5    cents  = 


271.  Aliquot  Parts  of  a  Pound  Sterling. 


10  shil.  =  ££• 

6s.  8d.  =  £:V 

5  shil.  -&\ 

4  shil.  =£} 


33.  4d.  = 
2s.  6d.  = 
2  shil.  = 
is.  8d.  = 


272.  Aliquot  Parts  of  a  Pound  Avoirdupois. 

pound. 


12  ounces  =  J  pound. 

8   -   =i   « 

si  "   =4  " 


4  ounces 
2      " 
i      " 


273.  Aliquot  Parts  of  a  Year. 


9  months  =  5  year. 
8      "         =5    " 
6      "         =i     " 
4      "         =4    " 


3  months  =  i  year. 

2         «  =          » 


274. 


o/"  a  Month. 


20  days  =  ^  month. 

is  »  =i   «• 

10   "    =4     " 
6    "    =4     " 


5  days  =    ^  month, 
3    "      =-Ar      " 

2      "       =^       « 


KEDUCTION. 

275.  Reduction  is  changing  a  number  from  one 
denomination  to  another,  without  altering  its  value.    It  is 
either  descending  or  ascending. 

Reduction  Descending  is  changing  higher  de- 
nominations to  lower ;  as,  yards  to  feet,  etc. 

Reduction  Ascendiny  is  changing  lower  denomi- 
nations to  higher ;  as,  feet  to  yards,  etc. 

276.  To  reduce  Higher  Denominations  to  Lower. 

1.  How  many  farthings  are  there  in  £23,  75.  5^d.  ? 
ANALYSIS. — Since  there  are  203.  in  a  OPERATION. 

pound,  there  must  be  20  times  as  many          £23,  78.  5d.    I  far. 

shillings  as  pounds,  plus  the  given  shil-  20 

lings.    Now  20  times   23   are  460,   and  "77  ~ 

4603.  +  7s.:=467S.     Again,  since  there  are 

I2d.   in    a   shilling,   there    must   be    12 

times  as  many  pence  as  shillings,  plus  the          5"°9  u. 

given  pence.    But  12  times  467  are  5604,     4 

and   5&04d. +  5d.  =  56oo.d.     Finally,   since        22437  ^'J[X'  Ans. 
there  are  4  farthings  in  a  penny,  there 

must  be  4  times  as  many  farthings  as  pence,  plus  the  given  farthings. 
Now  4  times  5609  are  22436,  and  22436  far. +  i  far.=22437  tar 
Therefore,  in  £23,  73.  s^d.  there  are  22437  far.  Hence,  the 

RULE. — Multiply  the  highest  denomination  by  the  number 
required  of  the  next  lower  to  make  a  unit  of  the  higher,  and 
to  the  product  add  the  lower  denomination. 

Proceed  in  this  manner  with  the  successive  denomina- 
tions, till  the  one  required  is  reached. 

2.  How  many  pence  in  £8,  IDS.  7d.  ?         Ans.  2047d. 

3.  How  many  farthings  in  123.  pd.  2  far.? 

4.  How  miny  farthings  in  £41,  53.  4^d.  ? 

27-;.  What  is  Reduction?  TIow  mrtnv  kinds?  Descending?  Asrenrtin"? 
276.  Hov  nre  higher  denomination*  reduced  to  lower?  Explain  Es.  i  from  th« 
blackboard 't 


190  SEDUCTION. 

277.  To  reduce  Lower  Denominations  to  Higher. 

5.  In   22437  farthings,  how  many  pounds,   shillings, 
pence,  and  farthings  ? 

ANALYSIS. — Since  iu  4  farthings  there  OPERATION. 

is  I  penny,  in  22437  farthings  there  are  4)22437  far. 

as  many  pence  as  4  farthings  are  con-  i2^^6oQd    I  far 

tained    times    in    22437    farthings,    or 
5609  pence  and  i  farthing  over.     Again, 

since  in  12  pence  there  is  I  shilling,  in  <£23>  7& 

5609  pence  theru  are  as  many  shillings  AllS.  £23,  78.  5d.  I  far. 
as  12  pence  are  contained  times  in 

5609  pence,  or  467  shillings  and  5  pence  over.  Finally,  since  in 
20  shillings  there  is  i  pound,  in  467  shillings  there  are  as  many 
pounds  as  20  shillings  are  contained  times  in  467  shillings,  or 
23  pounds  and  7  shillings  over.  Therefore,  in  22437  farthings  there 
are  £23,  73.  sd.  i  far.  Hence,  the 

RULE. — Divide  the  given  denomination  by  the  number  re- 
quired of  this  denomination  to  make  a  unit  of  the  next  higher. 

Proceed  in  this  manner  with  the  successive  denomina- 
tions, till  the  one  required  is  reached.  The  last  quotient) 
with  the  several  remainders  annexed,  will  be  the  answer. 

NOTE. — The  remainders,  it  should  be  observed,  are  the  same 
denomination  as  the  respective  dividends  from  which  they  arise. 

278.  PROOF. — Reduction   Ascending   and  Descending 
prove  each  other ;  for,  one  is  the  reverse  of  the  other. 

6.  In  2047  pence  how  many  pounds,  shillings,  and  pence  ? 

7.  In  614  farthings,  how  many  shillings,  pence,  etc.? 

8.  How  many  pounds,  shillings,  etc.,  in  39610  farthings  ? 

9.  An  importer  paid  £27,  135.  8d.  duty  on  a  package  of 
English   ginghams,  which  was  4d.  a   yard:   how  many 
yards  did  the  package  contain  ? 

10.  A  railroad  company  employs  1000  men,  paying  each 
43.  6d.  per  day :  what  is  the  daily  pay-roll  of  the  company  ? 

11.  Reduce  18  Ibs.  6  ounces  troy,  to  pennyweights. 


277.  How  are  lower  denominations  reduced  to  higher  ?    27".  Proof?    Explain 
Ex.  5  from  the  blackboard. 


REDUCTION.  191 

12.  Reduce  32  Ibs.  9  oz.  5  pwt.  to  pennyweights. 

13.  How  many  pounds  and  ounces  in  967  ounces  troy? 

14.  How  many  pounds,  etc.,  in  41250  grains? 

15.  How  many  rings,  each  weighing  3  pwt.,  can    bo 
made  of  i  Ib.  10  oz.  9  pwt.  of  gold  ? 

1 6.  What  is  the  worth  of  a  silver  cup  weighing  10  oz. 
16  pwt.,  at  \z\  cents  a  pennyweight? 

17.  Reduce  165  Ibs.  13  oz.  Avoirdupois  to  ounces. 

18.  Reduce  210  tons  121  pounds  8  ounces  to  ounces. 

19.  In  4725  Ibs.  how  many  tons  and  pounds? 

20.  In  370268  ounces,  how  many  tons,  etc.  ? 

21.  What  will  875  Ibs.  1 1  oz.  of  snuff  come  to,  at  5  cents 
an  ounce  ? 

22.  A  grocer  bought  i  ton  of  maple  sugar,  at  10  cents 
a  pound,  and  sold  it  at  6  cents  a  cake,  each  cake  weighing 
4  oz. :  what  was  his  profit  ? 

23.  In  250  Ibs.  6  oz.  Apothecaries'  weight,  how  many 
drams  ? 

24.  In  2165  scruples,  how  many  pounds  ? 

25.  How  many  feet  in  1250  rods,  4  yards,  and  2  feet? 
For  the  method  of  multiplying  by  5^  or  16^,  see  Art.  165. 

Ans.  20639  feet. 

26.  How  many  feet  in  365  miles  78  rods  and  8  feet? 

27.  In  5242  feet,  how  many  rods,  yards  and  feet? 

REMARK. — In  order  to  divide  the  second        5)1:242  ft. 

dividend  1747  yards  by  5.^,  the  number         x, -•       ,., 

of  yards  in  a  rod,' we  reduce  both  the      5?JI747  J   •  l    k 

divisor    and    dividend   to  halves;    then      ' 

divide  one  by  the  other.     Thus,  5^=n       I  03494  half  yds, 
halves  ;  and  1747=3494  halves ;  and  n  is    Ans.  317  r.  3^  yd.  I  ft. 
contained  in  3404,  317  times  and  7  over. 

But  the  remainder  7,  is  halves;  for  the  dividend  was  halves;  and 
7  half  yards=3^  yards.    (Art.  168,  a,  160.) 

28.  Reduce  38265  feet  to  miles,  etc. 

29.  How  many  rods  in  461  leagues,  2  miles,  6  furlongs? 


192  EEDUCTIOJT. 

30.  Reduce  7  m.  6  fur.  23  r.  5  yds.  8  in.  to  inches,  and 
prove  the  operation. 

31.  Eeduce  i  in.  7  fur.  39  r.  5  yds.  i  ft.  6  in.  to  inches, 
'  and  prove  the  operation. 

32.  If  a  farm  is  2\  miles  in  circumference,  what  will  it 
cost  to  enclose  it  with  a  stone  wall,  at  $1.85  a  rod? 

33.  At  6 \  cents  a  mile,  what  will  be  the  cost  of  a  trip 
round  the  world,  allowing  it  to  be  8299.2  leagues  ? 

34.  How  many  eighths  of  a  yard  in  a  piece  of  cloth 
5  7  yards  long  ? 

35.  How  many  sixteenths  of  a  yard  in  163  yards  ? 

36.  In  578  fourths,  how  many  yards? 

37.  In  1978  sixteenths,  how  many  yards? 

38.  How  many  vests  will  i6|  yards  of  satin  make,  allow- 
ing \  yard  to  a  vest  ? 

39.  A  shopkeeper  paid  $2.50  for  i8|  yards  ribbon,  and 
making  it  into  temperance  badges  of  -J-  yard  each,  sold 
them  at  \2\  cents  apiece:  what  was  his  profit? 

40.  How  many  square  rods  in  43816  square  feet? 

41.  How  many  acres,  etc.,  in  25430  square  yards  ? 

42.  Reduce  160  acres,  25  sq.  rods,  8  sq.  ft.  to  sq.  feet. 

43.  Reduce  too  sq.  miles  to  square  rods? 

44.  A  man  having  64  acres  8  sq.  rods  of  land,  divided 
it  into  6  equal  pastures:  how  much  land  was  there  in 
each  pasture  ? 

45.  My  neighbor  bought  a  tract  of  land  containing  15^ 
acres,  at  $500  an  acre,  and  dividing  it  into  building  lots 
of  20  square  rods  each,  sold  them  at  $250  apiece :  how 
much  did  he  make  ? 

46.  Reduce  85  cubic  yards,  10  cu.  feet  to  en.  inches. 

47.  Reduce  250  cords  of  wood  to  cu.  feet. 

48.  Reduce  18265  cu-  inches  to  cu.  feet,  etc. 

49.  Reduce  8278  cu.  feet  to  cords. 

50.  Reduce  164  bu.  i  pk.  3  qts.  to  quarts. 

51.  Reduce  375  bu.  3  pks.  i  qt.  i  pt.  to  pints. 


REDUCTION.  193 

52.  Reduce  184  chaldrons  17  bu.  to  bushels. 

53.  In  8147  quarts  how  many  bushels,  etc.  ? 

54.  A  seedsman  retailed  75  bu.  3  pk.  of  clover  seed  at 
17  cents  a  quart :  what  did  it  come  to  ? 

55.  A  lad  sold  2  bu.  i  pk.  3  qts.  of  chestnuts,  at  8£  cents 
a  pint:  how  much  did  he  get  for  them  ? 

56.  In  98  quarts  i  pt.  2  gills,  how  many  gills? 

57.  In  150  gals.  3  qts.,  how  many  quarts? 

58.  In  45  barrels,  10  gals.  3  qts.,  how  many  quarts? 

59.  How  many  gills  in  17  hhds.  10  gals.  3  qts.  ? 

60.  How  many  gallons,  etc.,  in  86673  pints  ? 

6 1.  How  many  bottles  holding  i£  pt.  each  are  required 
to  hold  a  barrel  of  wine  ? 

62.  What  vvill  a  hogshead  of  alcohol  come  to,  at  6£  cents 
a  gill? 

63.  Reduce  30  days  5  hours  42  min.  10  sec.  to  seconds. 

64.  Reduce  17  weeks  3  days  5  hr.  30  min.  to  minutes. 

65.  Reduce  a  solar  year  to  seconds. 

66.  Reduce  6034500  sec.  to  weeks,  etc. 

67.  Reduce  5603045  hours  to  common  years. 

68.  Reduce  10250300  minutes  to  leap  years. 

69.  An  artist  charged  me  75  cents  an  hour  for  copying 
a  picture;  it  took  him  27  days,  working  io£  hours  a  day: 
wliat  was  his  bill  ? 

70.  In  48561  seconds,  how  many  degrees? 

71.  In  65237  minutes,  how  many  signs? 

72.  In  237°,  40',  31",  how  many  seconds? 

73.  In  360°,  how  many  seconds? 

74.  How  many  dozen  in  1965  things? 

75.  How  many  eggs  in  125  dozen? 

76.  How  many  gross  in  100000? 

77.  How  many  pens  in  65  gross? 

78.  How  many  pounds  in  3  score  and  7  pounds? 

79.  How  many  sheets  of  paper  in  75  quires? 
So.  How  many  quires  of  paper  in  i  oooo  sheets  ? 


194 


APPLICATIONS     OF 


APPLICATIONS   OF  WEIGHTS   AND    MEASURES. 


4  yards. 


279.  The  use  of  Weights  and  Measures  is  not 

confined  to  ordinary  trade.  They  have  other  and  im- 
portant applications  to  the  farm,  the  garden,  artificers' 
work,  the  household,  etc. 

THE     FARM    AND     GARDEN. 

280.  To  find  the  Contents  of  Rectangular  Surfaces. 

1.  How  many  square  yards  in  a  strawberry  bed,  4  yards 
long,  and  3  yards  wide  ? 

ILLUSTRATION. — Let  the  bed  be  rep- 
resented by  the  adjoining  figure;  its 
length  being  divided  into  four  equal    . 
parts,  and  its  breadth  into  three,  each  'g 
denoting  linear  yards.     The  bed,  ob-  >> 
viously,  contains  as  many  square  yards 
RB    them   are   squares  in   the    figure. 
Now  as  there  are  4  squares  in  i  row,  in 
3  rows  there  must  be  3  times  4  or  12 
squares.     Therefore,  the  bed  contains  12  sq.  yards.     Hencj,  the 

RULE. — Multiply  the  length  by  the  breadth.     (Art.  243.) 

NOTES. — i.    Both   dimensions   should   be   reduced  to  the   sam-e 
denomination  before  they  are  multiplied. 

2.  The  area  and  one  side  of  a  rectangular  surface  being  given,  the 
other  side  is  found  by  dividing  the  area  by  the  given  side. 

3.  One  line  is  said  to  be  multiplied  by  another,  when  the  number 
of  units  in  the  former  are  taken  as  many  times  as  there  are  like  u n  its 
in  the  latter.    (Art.  47,  n.) 

2.  How  many  acres  in  afield  40  rd.  long  and  31  rd.  wide  ? 

wide? 

3.  If  the  width  of  an  asparagus  bed  is  66  feet,  what 
must  be  the  length  to  contain  \  of  an  acre  ? 


279.  What  is  sairt  of  the  application*  of  Weight*  and  Measures?  280.  How 
fin  1  the  area  of  a  rectangular  surface  1  Note.  If  t;.e  dimensions  are  in  different 
denominations,  what  is 'to  be  done  ?  If  the  area  and  one  side  are  given  ? 


WEIGHTS     AND     MEASURES.  19fi 

4.  What  is  the  length  of  a  pasture,  containing  15  acres, 
whose  width  is  33  J  rods  ? 

5.  A  speculator  bought  30   acres  of  land  at  $50  per 
acre,  and  sold  it  in  villa  lots  of  5  rods  by  4  rods,  at  $200  a 
lot:  what  was  his  profit? 

6.  A  garden  230  ft.  long  and  125  ft.  wide,  has  a  gravel 
walk  6  ft.  in  breadth  extending  around  it:  how  much 
land  does  the  walk  contain  ? 

7.  What  is  the  difference  between  2  feet  square  and 
2  square  feet  ? 

8.  A  farmer  having  3  acres  of  potatoes,  sold  them  at 
25  cents  a  bushel  in  the  ground ;  allowing  a  yield  of  2\  bu. 
to  a  sq.  rod,  how  many  bushels  did  the  field  produce ;  and 
what  did  he  receive  for  them  ? 

281.    To  find  the  number  of  Hills,  Vines,  etc.,    in  a  given 
field,  the  Area  occupied  by  each  being  given. 

9.  How  many  hills  of  corn  can  a  farmer  plant  on  2  acres 
of  ground,  the  hills  being  4  ft.  apart  ? 

ANALYSIS. — Since  the  hills  are  4  ft.  apart,  each  hill  will  occupy 
4  x  4  or  16  sq.  ft.  Now  2  acres=  160  sq.  rods  x  272.25  x  2=87120  sq.  ft. 
Therefore,  he  can  plant  as  many  hills  as  16  is  contained  times  ip 
87120;  and  87120-^  16=5445  hills.  Hence,  the 

RULE. — Divide  tJie  area  planted  by  ilw  area  occupied  ly 
each  hill  or  vine. 

10.  How  many  bulbs  will  a  lady  require  for  a  crocus 
bed,  1 2  ft.  long  and  4  ft.  wide,  if  planted  6  inches  apart  ? 

11.  What  number  of  grape  Vines  will  be  required  for 
$  of  an  acre,  if  they  are  set  3  ft.  apart  ?     What  will  it  cost 
to  set  them  at  $5.25  per  hundred  ? 

12.  How  large  an  orchard  shall  I  require  for  100  apple 
trees,  allowing  them  to  stand  33  ft.  apart  ? 

281.  Hew  find  the  number  of  hills,  vines,  etc.,  In  a  given  field?  284.  How  »n* 
Uooring,  plastering,  etc.,  estimated? 


196  A  !>  P  L I  C  A  1 1 0  IT  8     0  Jf 

ARTIFICERS'     WORK. 

282.  Flooring,  plastering,  painting,  papering,  roofing, 
paving,  etc.,  are  estimated  by  the  number  of  sq.ft.  or  sq.  yds. 
in  the  area,  or  by  the  "square"  of  100  sq.  feet. 

13.  What  will  be  the  cost  of  flooring  a  room  18  ft.  long 
and  1 6  ft.  6  in.  wide,  at  i8|  cts.  a  sq.  foot? 

SOLUTION. — 18  ft.  x  16^=297  sq.  ft. ;  and  i8J  cts.  x  297= $55.68 J. 

14.  If  a  school  house  is  60  ft.  long  and  45  ft.  wide,  what 
will  be  the  expense  of  the  flooring,  at  61.08  per  sq.  yard  ? 

15.  What  will  it  cost  to  plaster  a  ceiling  18  ft.  6  in.  long, 
and  15  ft.  wide,  at  83.20  per  square  of  100  sq.  ft, 

1 6.  What  will  be  the  cost  of  flagging  a  side-walk  206  ft 
long  and  9^  ft.  wide,  at  $1.85  per  sq.  yard  ? 

17.  What  will  be  the  expense  of  painting  a  roof  60  ft 
long  and  25  feet  wide,  at  $1.50  per  "square"  ? 

1 8.  What  cost  the  cementing  of  a  cellar  65  ft  by  25  ft, 
at  $.25  per  square  yard  ? 

MEASUREMENT  OF   LUMBER. 

283.  To  find  the   Contents   of  Boards,  the  length  and 

breadth  being  given. 

DBF. — A  Standard  Board,  in  commerce,  is  i  in.  thick.  Hence, 
A  Board  Foot  is  i  foot  long,  i  foot  wide,  aud  i  inch  thick. 
A  Cubic  Foot  is  12  board  feet.    Hence,  the 

EULE. — Multiply  the  length  in  feet  by  the  width  in  inches, 
and  divide  the  product  by  12;  the  result  will  be  board  feet. 

NOTES. — i.  If  a  board  tapers  regularly,  multiply  by  the  mean 
width,  which  is  ludf  the  sum  of  the  two  ends. 

2.  Shingles  are  estimated  by  the  thousand,  or  bundle.     They  are 
commonly  18  in.  long,  and  average  4  in.  wide.      It  is  customary  to 
allow  looo  to  a  square  of  100  feet. 

3.  The  contents  of  boards,  timber,  etc.,  were  formerly  computed 
by  duodecimals,  which  divides  t\\efoot  into  12  in.,  the  inch  into  12 
sec.,  etc. ;  but  this  method  is  seldom  used  at  the  present  day. 

283.  The  thickness  of  a  t-tumlard  board  ?    What  is  a  board  foot  T     A  cubic 
foot  ?    lJ<nv  find  the  contents  of  a  board  ?    If  the  board  is  taperiug,  how  ? 


WEIGHTS     A5JD     MEASURES.  10? 

19.  If  a  board  is  10  ft.  long,  and  i  ft.  3  in.  wide,  w,hat 
are  its  contents,  board  measure? 

SOLUTION. — iox  15  =  150;  and  150-4-12  =  12^  board  ft.,  Ans. 

20.  What  are  the  contents  of  a  board  13  ft.  long,  and 
i  ft.  5  in.  wide  ? 

SOLUTION. — 13  x  17=221 ;  and  221-5-12=18  fe  board  ft. 

21.  Required  the  contents  of  a  board  14  ft.  long,  i  ft 
4  in.  wide ;  and  its  value  at  i\  cts.  a  foot  ? 

22.  Required  the  contents  of  a  tapering  board  15  ft 
long,  14  in.  wide  at  one  end,  and  6  in.  at  the  other. 

23.  Required  the  contents  of  a  stock  of  9  boards  14  ft 
long,  and  i  ft.  2  in.  wide. 

24.  The  sides  of  a  roof  are  45  ft.  long  and  20  ft.  wide ; 
what  will  it  cost  to  shingle  both  sides,  at  $15.45  per  M., 
allowing  1000  shingles  to  a  square  ? 

284.  To  find  the  Cubical  Contents  of  Rectangular  Bodies, 

REM. — Round  Timber,  as  masts,  etc.,  is  estimated  in  cubic  feet. 
Hewn  Timber,  as  beams,  etc.,  either  in  board  or  cubic  feet. 
Sawed  Timber,  as  planks,  joists,  etc.,  in  board  feet. 

25.  How  many  cubic  feet  are  there  in  a  rectangular 
block  of  marble  4  ft.  long,  3  ft.  wide,  and  2  ft.  thick  ? 

4  feet. 

ILLUSTRATION.— Let  the  block  be 
represented  by  the  adjoining  diagram  ; 
its  length  being  divided  into  four 
equal  parts,  its  bread tli  three,  and  its 
thickness  into  two,  each  denoting  linear 
feet. 

In  the  upper  face  of  the  block,  there  are  3  times  4,  or  12  sq.  ft.  Now 
if  the  block  were  I  foot  thick,  it  would  evidently  contain  i  time  as 
many  cubic  feet  as  there  are  square  feet  in  its  upper  face  ;  and  I  time 
4  into  3=12  cubic  feet.  But  the  given  block  is  2  ft.  thick ;  therefore 
it  contains  2  times  4  into  3=24  cu.  ft.  Hence,  the 

RULE. — Multiply  tlio  length,  breadth,  and  thickness 
together.  (Art.  249.) 


284.  How  are  hewn  timber,  etc.,  es  imaied?   How  joists,  studs,  etc.  ?    285.  How 


198  APPLICATIONS   OF 

285.  To  find  the  Contents  of  Joists,  ic.,  2, 3, 4,  &c.,  in.  thick. 

RULE. — Multiply  the  width  l>y  such  a  part  of  the  length, 
as  the  thickness  is  of  12 ;  the  result  will  be  board  feet. 

NOTES. — i.  The  approximate  contents  of  round  titriber  or  lorjs  may 
be  found  by  multiplying  \  of  the  mean  circumference  by  itself,  and 
this  product  by  the  length. 

2.  Cubic  feet  are  reduced  to  Board  feet  by  multiplying  them  by  12. 
For,  I  foot  board  measure  is  12  inches  square,  and  i  in.  thick  ;  there- 
fore, 12  such  feet  make  i  cubic  foot.    Hence, 

If  one  of  the  dimensions  is  inches,  and  the  other  two  arc  feet,  the 
product  will  be  in  Board  feet. 

3.  To  estimate  hay. — 

A  ton  of  hay  upon  a  scaffold  measures  about  500  cu.  ft. ;  when  in 
a  mow,  400  cu.  ft. ;  and  in  well  settled  stacks,  10  cu.  yards. 

4.  To  find  the  weight  of  coal  in  bins. — 

A  ton  (2000  Ibs.)  of  Lehigh  white  ash,  egg  size,  measures  34  V  cu.  ft. 
A  ton  of  white  ash  Schuylkill,  "  "         35  cu.  ft. 

A  ton  of  pink,  gray  and  red  ash,  "  "         36  cu.  ft. 

26.  How  many  cu.  feet  in  a  stick  of  hewn  timber  20  ft. 
long,  i  ft.  3  in.  wide,  i  ft.  4  in.  thick  ?        Ans.  33^  en.  ft 

27.  How  many  board  feet  in  a  joist  18  ft.  long,  5  in. 
wide,  and  4  in.  thick  ?     How  many  cubic  feet  ? 

SOLUTION. — 4  is  £  of  12 ;  and  i  of  18  ft.  is  6  ft.  Now  5  x6=3o 
bd.  ft. ;  and  30-^-12=2^  cu.  ft. 

28.  What  cost  45  pieces  of  studding  n  ft.  long,  4  in. 
wide,  and  3  in.  thick,  at  3  cts.  a  board  foot  ? 

29.  At  45  cts.  a  cu.  foot,  what  will  be  the  cost  of  a  beam 
32  ft.  long,  i  ft.  6  in.  wide,  and  i  ft.  i  in.  thick  ? 

30.  What  is  the  worth  of  a  load  of  wood  8  ft  long,  4  ft 
wide,  and  5  ft.  high,  at  $3!  a  cord  ? 

31.  How  many  tons  of  white  ash,  e-s  Sch.  coal,  will  a  bin 
10  ft  long,  8  ft.  wide,  and  8  ft  high,  contain.    Ans.  i8f  T. 

find  the  contents  of  a  rectangular  body  ?    Note.  How  find  the  contents  of  round 
timber?    When  the  contents  and  two  of  the  dimensions  are  given,  how  flTirt  the 
other  dimension?      TTnw  reduce  cubic  feet  of  timber  to  beard  feet?      Why 
1.  <>:M!  iHiinoihion  is  inches,  and  the  other  two  ft.,  what  is  the  product  T 


WEIGHTS     AXD     MEASURES.  199 

32.  How  many  cords  of  wood  in  a  tree  60  feet  high, 
whose  mean  circumference  is  10  feet? 

33.  Hov  many  tons  of  hay  in  a  mow  22  ft.  by  20  ft, 
and  15  ft.  high  ? 

34.  At  818.50  a  ton,  what  is  the  worth  of  a  scaffold  of 
hay  30  ft.  long,  12  ft.  wide,  and  10  ft.  high? 

286.  Stone  masonry  is  commonly  estimated  by  the 
perch  of  25  cubic  feet 

Excavations  and  embankments  are  estimated  by  the 
cubic  yard.  In  removing  earth,  a  cu.  yard  is  called  a  load. 

Brickwork  is  generally  estimated  by  the  1000,  but 
sometimes  by  cubic  feet 

NOTES. — i.  A  perch  strictly  8peaking=24|  cu.  feet  being  16^  ft. 
long,  ii  ft.  wide,  and  i  ft.  high. 

2.  The  average  size  of  bricks  is  8  in.  long,  4  in.  wide,  and  2  in.  thick. 

In  estimating  the  labor  of  brickwork  by  cu.  feet,  it  is  customary  to 
measure  the  length  of  each  wall  on  the  outside ;  BO  allowance  is 
made  for  windows  and  doors  or  for  corners. 

In  finding  the  exact  number  of  bricks  in  a  building,  we  should 
make  a  deduction  for  the  windows,  doors,  and  corners  ;  also  an  allow- 
ance of  -,'u  of  the  solid  contents  for  the  space  occupied  by  the  mortar. 

35.  What  will  be  the  cost  of  digging  a  cellar  45  feet 
long,  24  feet  wide,  and  8  feet  deep,  at  35  cents  a  cu.  yard  ? 

36.  At  $5.25  a  perch  (25  cu.  ft),  what   cost  the  labor 
of  building  a  cellar  wall,  i  ft.  6  in.  thick,  and  8  ft  high, 
the  cellar  being  22  by  45  feet? 

37.  How  many  bricks  of  average  size  will  it  take  to 
build  the  walls  of  a  house  50  ft  long,  35  ft  wide,  21  ft. 
high,  and  i  ft.  thick,  deducting  -^  of  the  contents  for  the 
space  occupied  by  the  mortar,  but  making  no  allowance 
for  windows,  doors,  or  corners  ? 

38.  At  45  cents  a  cubic  yard,  what  will  it  cost  to  fill  in 
a  street  800  ft  long,  60  ft.  wide,  and  4^  ft.  below  grade  ? 

286.  How  are  stone  masonry,  excavations,  etc.,  vthjamci  T  How  brir.U  work? 
tfot*.  The  overage  eixfl  of  bricks  ? 


200  APPLICATIONS     OK 

THE    HOUSEHOLD. 

287.  To  find  the  Quantity  of  material  of  given  width  required 
to  line  or  cover  a  given  Surface. 

39.  How  many  yards  of  serge  £  yd.  wide  are  required  to 
line  a  cloak  containing  7$  yds.  of  camlet  i-J  yd.  wide  ? 

ANALYSIS. — The  surface  to  be  lined=7J  yds.  x  i\=^  x  £  — Lf£  sq. 
yards  :  i  yd.  of  the  liniug=i  yd.  x  f =£  sq.  yd.  Now  as  f  sq.  yard 
of  camlet  requires  i  yard  of  lining,  the  whole  cloak  will  require  aa 
many  yards  of  lining  as  f  is  contained  times  in  J,56i;  and  iftf-t-^ 
Lr^  x  2=^-  or  15^  yards,  the  serge  required.  Hence,  the 

RULE. — Divide  the  number  of  square  yards  or  feet  in  the 
given  surface  by  the  number  of  square  yards  or  feet  in  a 
yard  of  the  material  used. 

40.  How  many  yards  of  Brussels  carpeting  £  yd.  wide, 
are  required  for  a  parlor  floor  21  ft.  long  and  18  ft.  wide? 

41.  How  many  yards  of  matting  1 J  yd.  wide  will  it  take 
to  cover  a  hall  16  yds.  long  and  3-^  yds.  wide  ? 

42.  How  much  cambric  -J  yd.  wide  is  required  to  line  a 
dress  containing  15  yds.  of  silk  £  yd.  wide  ? 

43.  How  many  sods,  each  being  15  in.  square,  will  be 
required  to  turf  a  court  yard  25  by  20  feet? 

44.  How  many  yards  of  silk  £  yd.  wide  are  required  to 
line  2  sets  of  curtains,  each  set  containing  10   yds.  of 
brocatelle  i-J  yd.  wide? 

45.  How  many  marble  tiles  9  in.  square,  will  it  take  to 
cover  a  hall  floor  48  ft.  long  and  7^  ft.  wide  ? 

46.  How  many  rolls  of  wall  paper  9  yds.  long  and  i£  ft. 
wide,  are  required  to  cover  the  4  sides  of  a  room  18  by  16  ft. 
and  9  ft.  high,  deducting  81  sq.  ft.  for  windows  and  doors? 

47.  How  many  gallons  of  water  in  a  rectangular  cistern, 
whose  length  is  8  ft,  breadth  6  ft.,  and  depth  5  ft.  ? 

NOTE. — Cubic  measure  is  changed  to  gallons  or  bushels,  by  reducing 
the  former  to  cu.  inches,  and  diciding  the  result  by  231,  or  2150.4,  aa 
the  case  may  be.  (Arts.  252,  255.) 


*8/.  How  find  the  quantitj  of  material  required  to  line  or  cover  a  given  purfaca ! 


WEIGHTS     AND     MEASURES. 

48.  A  man  constructed  a  cistern  containing  20  hogs- 
heads, the  base  being  6  ft-  square :  what  was  its  depth  ? 

49.  How  many  bushels  of  wheat  will  a  bin  6  ft.  long, 
\  ft.  wide,  and  3  ft.  deep,  contain  ? 

50.  How  many  cu.  feet  in  a  vat  containing  50  hogsheads  ? 

NOTE. — Gallons  and  bushels  are  reduced  to  cubic  inches  by  multi- 
plying them  by  231  or  2150.4,  as  the  case  may  be. 

51.  A  man  having  10  bu.  of  grain,  wishes  to  store  it  in 
a  box  5  ft.  long  and  3  ft.  wide :  what  must  be  its  height  ? 

52.  A  man  built  a  cistern  in  his  attic,  5  ft.  long,  4  ft. 
wide,  and  3  ft.  deep :  what  weight  of  water  will  it  contain, 
allowing  1000  oz.  to  a  cu.  ft.  ? 

288.    To  change   Avoirdupois    Weight  to   Trey,  and  Troy 
Weight  to  Avoirdupois. 

53.  If  the  internal  revenue  tax  is  5  cents  per  ounce 
Troy,  on  silver  plate  exceeding  40  ounces,  what  will  be  the 
tax  on  plate  weighing  7  Ib.  8  oz.  Avoirdupois  ? 

ANALYSIS. — 7  Ib.  S  oz.  Avoirdupois =120  oz. :  and  120  oz.  X437^  = 
52500  grains.  Now  52500  gr.-f- 480=:  109!  oz.  Troy.  Again,  109}  oz. 
—40  oz.  (exempt)=69|  oz. ;  and  5  cts.  x  6gf:=$3.461.  Ans. 

54.  How  many  spoons,  each  weighing  i  oz.  Avoirdupois, 
can  be  made  from  i  Ib.  9  oz.  17  pwt.  12  gr.  Troy? 

ANALYSIS. — i  Ib.  9  oz.  17  pwt.  12  gr.= 10500  grains;  and  10500 
£rs.-t-437.5  grs.  (i  oz.  Avoir.)=24  spoons.  Ans.  Hence,  the 

RULE. — Reduce  the  given  quantity  to  grains;  then  reduce 
t)w  grains  to  the  denominations  required.  (Art.  231.) 

55.  If  a  family  have  1 2  Ib.  6  oz.  Avoir,  of  silver  plate,  what 
will  be  the  tax,  exempting  40  oz.,  at  5  cts.  per  oz.  Troy  ? 

56.  A  lady  bought  a  silver  set  weighing  n  Ibs.  4  oz. 
Troy  :  how  many  pounds  Avoir,  should  it  weigh  ? 

57.  How  many  rings,  each  weighing  3^  pwt.,  can  be 
made  from  a  .bar  of  gold  weighing  i  pound  Avoirdupois? 

58.  What  is  the  value  of  a  silver  pitcher  weighing  2  Ib. 
8  oz.  Avoir.,  at  $2  per  ounce  Troy? 

59.  A  miner  sold  a  nugget  of  gold  weighing  3$  Iba 
Avoir.,  at  $17  per  oz.  Troy :  what  did  he  get  for  it  ? 


DENOMINATE    FRACTIONS. 

289.  A  Denominate  Fraction,  is  one  or  more  of 
the  equal  parts  into  which  a  compound  or  denominate 
number  is  divided. 

Denominate  Fractions  are  expressed  either  v.r 
common  fractions,  or  as  decimals.  The  former  are  usually 
termed  denominate  fractions;  the  latter,  denominate 
decimals. 

REDUCTION    OF   DENOMINATE   FRACTIONS. 
CASE   I. 

290.  To   reduce  a   Denominate   Fraction   from   Higher 

denominations  to  Lower. 

1.  What  part  of  an  inch  is  -fe  of  a  yard  ? 

ANALYSIS.  —  ^  of  i  yard  equals  ^  of  2  yards.  2  yd.  x  3  =  6  ft. 
Reducing  the  numerator  to  inches,  we  have  6  ft.  X  1  2  —  7  2  in. 
2  yards-  2  x  3  x  12,  or  72  inches  ;  and  9^  of  72  ^ns.  £*  or  4  in. 
in.  is  H  or  £  inch. 

Or,  denoting  the  multiplications,  and  cancelling  the  factors 
common  to  the  numerators  and  denominators,  we  have 

2  2  X  •?  .        2X3X12.          2  X  •$  X  12      ?. 

—  yd.  =  -  2.  ft  =  -  !L—  m.  =  —  2.  —  ~_  =  3  in.    Hence,  the 
96  96  96  96,  4        4 

RULE.  —  Reduce  the  numerator  to  the  denomination  re- 
quired, and  place  it  over  the  given  denominator;  cancdliity 
the  factors  common  to  both.  (Art.  276.) 

NOTE.  —  The  steps  in  this  operation  are  the  same  as  those  in 
Reduction  Descending. 

2.  Reduce  ^  of  a  bushel  to  the  fraction  of  a  quart. 

3.  Reduce  £^5-  to  the  fraction  of  a  penny. 

4.  Reduce  -^^  of  a  day  to  the  fraction  of  an  hour. 

5.  What  part  of  an  ounce  is  -^  of  a  pound  ? 

6.  What  part  of  a  square  inch  is  -yfj  of  a  sq.  foot  ? 


289.  What  are  denominate  fractions?  How  are  they  expressed?  290.  Row 
ttr«  denominate  fractions  reduced  from  hijrher  denominations  to  lower?  Note 
What  are  the  etepn  in  this  operation  like?  Explain  Ex,  t  from  the  blackboard. 


DENOMINATE     FB  ACTIONS. 

CASE    II. 

293.    To   reduce  a   Denominate   Fraction  from   Lower 
denominations  to  Higher. 

7.  What  part  of  a  yard  is  |  of  an  inch  ? 

ANALYSIS. — Reducing  I  yard  to  fourths  of  an  inch,  we  have  I  yd. 
=  36  in. ;  and  36  in.  x  4=144  fourths  inch.  Now  3  fourths  are  Tf  3  ol 
144  fourths ;  therefore  f  of  an  inch  is  T^T  or  41s  of  a  yard.  Hence,  the 

RULE. — Reduce  a  unit  of  the  denomination  in  which  the 
required  fraction  is  to  be  expressed,  to  the  same  denomination 
as  the  given  fraction,  and  place  the  numerator  over  it. 

8.  What  part  of  a  dollar  is  £  of  a  mill  ? 

9.  What  part  of  a  pound  is  £  of  a  farthing  ? 
10.  What  part  of  a  Troy  ounce  is  -^  of  a  grain  ? 
n.  Reduce  f  of  a  gill  to  the  fraction  of  a  gallon. 

12.  Reduce  £  of  a  rod  to  the  fraction  of  a  mile. 

1 3.  Reduce  £  of  a  pound  to  the  fraction  of  a  ton. 

CASE    III. 

292.    To    reduce    a   Denominate  Fraction  from    higher 
to  a  IVliole  Number  of  lower  denominations. 

14.  Reduce  f  of  a  yard  to  feet  and  inches. 

ANALYSIS. — Reducing  the  given  fraction  to  the  next  lower  denomi- 
nation, we  have  £  yard  x  3=16i  ft.  or  2  ft.  and  I  ft.  remainder. 

Again,  reducing  the  remainder  to  the  next  lower  denomination, 
we  have  £  ft.  x  12= V-» or  4a  iQ-  Therefore,  £  yards  equals  2  ft.  4^  in. 
(Arts.  147,  276.)  Hence,  the 

RULE. — Multiply  the  given  numerator  by  the  number 
required  to  reduce  the  fraction  to  the  ne.it  lower  denomina- 
tion, and  divide  the  product  by  the  denominator.  (Art.  276.) 

Multiply  and  divide  the  successive  remainders  in  the 
same  manner  till  the  lowest  denomination  is  reached.  The 
several  quotients  will  be  the  answer  required. 


291.  now  are  denominate  fractions  reduced  from  lower  denominations  to 
higher?  2^2.  How  arc  denominate  fractions  reducwl  from  higher  to  wLoie 
numbers  of  lower  denominations  T 


204  DENOMINATE     FRACTIONS. 

15.  Eeduce  £  of  an  eagle  to  dollars  and  cents.  Ans.  $7.50 

16.  Reduce  f  of  a  pound  sterling  to  shillings  and  pence 

17.  How  many  pecks  and  quarts  in  -ff  of  a  bushel  ? 

1 8.  How  many  pounds  in  f  of  a  ton  ? 

19.  How  many  ounces,  etc.,  in  |  of  a  Troy  pound  ? 

20.  What  is  the  value  of  -^  of  a  mile  ? 

21.  What  is  the  value  of  f|  of  an  acre  ? 

22.  Keduce  |  of  a  cord  to  cu.  feet 

CASE   IV. 

293.    To  reduce  a   Compound  Number  from  lower  to  a 
Denominate  Fraction  of  higher  denominations. 

23.  Reduce  2  ft.  3  in.  to  the  fraction  of  a  yard. 
ANALYSIS. — 2  ft.  3  in.=27  inches;  and  i  yard=36  inches.     The 

question  now  is,  what  part  of  36  in.  is  27  in.  ?    But  27  is  f  J  or  J  of  36. 
(Art.  173.)     Therefore  2  ft.  3  in.  is  ^  of  a  yard.     Hence,  tho 

RULE. — I.  Reduce  the  given  number  to  its  lowest  denom- 
ination for  the  numerator. 

II.  Reduce  to  the  same  denomination,  a  unit  of  the 
required  fraction,  for  the  denominator. 

NOTES. — i.  If  the  lowest  denomination  of  the  given  number  con- 
tains a  fraction,  the  number  must  be  reduced  to  the  parts  indicated 
by  the  denominator  of  the  fraction. 

2.  If  it  is  required  to  find  ichnt  part  one  compound  number  is  of 
another  which  contains  different  denominations,  reduce  both  to  tht 
lowest  denomination  mentioned  in  either,  and  proceed  as  above. 

24.  Reduce  2  qt.  i  pt.  3!  gills  to  the  fraction  of  a  gal.  ? 

25.  What  part  of  a  pound  Troy  is  5  oz.  2  pwt.  ic  gr.  ? 

26.  Reduce  3  pk.  2  qt.  i  pt.  to  the  fraction  of  a  bushel 

27.  Reduce  6  cwt.  48  Ib.  to  the  fraction  of  a  ton. 

28.  What  part  of  a  square  yard  is  5  sq.  ft.  2of  sq.  inches  ? 

29.  What  part  of  £3,  IDS.  2d.  is  75.  9d.  2  far.  ? 

30.  What  part  of  a  week  is  3  days,  5  hrs.  40  min  ? 

3 1.  What  part  of  a  cord  is  25^  cu.  feet  of  wood  ? 

32.  What  part  of  5  miles  2  fur.  14 rods  is  2  m.  i  fur.  2  rods? 


293.  How  arc  denominate  numbers  reduced  from  lower  denominations  to 
fractions  of  a  higher  denomination  ?    Explain  Ex.  23  from  tho  blackboard. 


DENOMINATE     DECIMALS.  205 


CASE    V. 

294.  To  reduce  a  Denominate  Decimal  from  a  higher  to  a 
Compound  Number  of  lower  denominations. 

33.  Reduce  .89375  gallon  to  quarts,  pints,  and  gills? 

ANALYSIS. — This  example  is  a  case  of 

Reduction  Descending.    (Art.  276.)    We  '89375  gal. 

therefore  multiply  the  given  decimal  of  a  4 

gallon  by  4  to  reduce  it  to  the  next  lower  3.5 7500  qt. 

denomination,  and  pointing  off  the  pro-  2 

duct  as  in  multiplication  of  decimals,  the  , 

result  is  3  qts.  and  .57500  qt.     In   like 
manner,  we  multiply  the  decimal  .57500  -   . 

qt.  by  2  to  reduce  it  to  pints,  and  the  .60006  gi. 

result  is  i  pt.  and  .15000  pt.     Finally,  we     Ans.  3  qt.  I  pt.  0.6  gi. 
multiply  the  decimal  .15000  pt.  by  4  to 

reduce  it  to  gills,  and  the  result  is  .6  gill.     Therefore,  .89375  gal. 
equals  3  qt.  i  pt.  0.6  gi.     Hence,  the 

RULE. — Multiply  the  denominate  decimal  by  the  number 
required  of  the  next  lower  denomination  to  make  one  of  the 
given  denomination,  and  point  off  the  product  as  in  multi' 
plication  of  decimals.  (Arts.  191,  276.) 

Proceed  in  this  manner  with  the  decimal  part  of  the  suc- 
cessive products,  as  far  as  required.  The  integral  part  of 
the  several  products  will  be  the  answer. 

34.  What  is  the  value  of  £.125445  in  shillings,  etc.  ? 

35.  What  is  the  value  of  .91225  of  a  Troy  pound? 

36.  Reduce  .35  mile  to  rods,  etc. 

37.  How  many  quarts  and  pints  in  .625  of  a  gallon? 

38.  How  many  minutes  and  seconds  in  .65 1  degree  ? 

39.  How  many  days,  hours,  etc.,  in  .241  week  ? 

40.  What  is  the  value  of  .25256  ton  ? 

41.  What  is  the  value  of  .003  of  a  Troy  pound  ? 

42.  What  is  the  value  of  £5.62542  ? 

294.  How  are  denominate  decimals  reduced  from  higher  denominations  to 
whole  numbers  of  a  lower  denomination  ?  Explain  Ez.  33  upon  the  blackboard 


206  DENOMINATE     DECIMALS. 

CASE    VI. 

295.    To  reduce  a  Compound  Number  from  lower  to  a 
Denominate  Decimal  of  a  higher  denomination. 

43.  Reduce  3  pk.  2  qt.  i  pt.  to  the  decimal  of  a  bushel. 

ANALYSIS. — 2  pints  make  i  quart ;  hence  OPERATION. 

there  is  \  as  many  quarts  as  pints,  and  £  of  21  pt. 

i  qt.  is  .5  qt.,  which  we  write  as  a  decimal  on  o      c  of 

the  right  of  the  given  quarts.    In  like  manner  _1 

there  is  £  as  many  pecks  as  quarts,  and  \  of 


2.5  is  .3125  pk.,  which  we  place  on  the  right     Ans.  .828125  bu. 
of  the  given   pecks.     Finally,  there  is  |  as 

many  bushels  as  pecks,  and  ^  of  3.3125  is  .828125  bu.     Therefore, 
3  pk.  2  qt.  i  pt.  equals  .828125  bushel.     Hence,  the 

RULE. —  Write  the  numbers  in  a  column,  placing  the 
lowest  denomination  at  the  top. 

Beginning  with  the  lowest,  divide  it  by  the  number  re- 
quired of  this  denomination  to  make  a  unit  of  the  next 
higher,  and  annex  the  quotient  to  the  next  denomination. 

Proceed  in  this  manner  with  the  successive  denomina- 
tions, till  the  one  required  is  reached. 

44.  Reduce  6  oz.  10  pwt.  4  gr.  to  the  decimal  of  a  pound. 

45.  Reduce  10  Ib.  to  the  decimal  of  a  ton. 

46.  Reduce  3  fur.  25  r.  4  yd.  to  the  decimal  of  a  mile. 

47.  Reduce  9.6  pwt.  to  the  decimal  of  a  Troy  pound. 

48.  What  decimal  part  of  a  barrel  are  15  gal.  3  qt. 

49.  What  decimal  part  of  2  rods  are  2\  fathoms  ? 

50.  What  decimal  part  of  £3,  33.  6d.  are  158.  io.$d.  ? 

51.  What  part  of  a  barrel  are  94.08  Ibs.  of  flour? 

52.  Reduce  7.92  yds.  to  the  decimal  of  a  rod. 

53.  Reduce  i  day  4  hr.  10  sec.  to  the  decimal  of  a  week. 

54.  Reduce  45  sq.  rods  25  sq.  ft.  to  the  decimal  of  an  acre. 

55.  Reduce  53^  cu.  feet  to  the  decimal  of  a  cord. 


295.   flow  arc  compound  numbers  reduced   from    lower   denominations  tf 
denominate  decimals  of  a  higher?    Explain  Ex.  43  upon  the  blackboard? 


METEIO    WEIGHTS 
MEASUEES. 

296.  Metric  Weight  sand  Measures  are  founded 
upon  the  decimal  notation,  and  are  so  called  because  their 
primary  unit  or  lase  is  the  Meter.* 

297.  The  Meter  is  the  unit  of  length,  and  is  equal  to 
one  ten-millionth  part  of  the  distance  on  the  earth's  surface 
from  the  equator  to  the  pole,  or  39.37  inches  nearly. 

NOTES.  —  i.  The  term  meter  is  from  the  Greek  metron,  a  measure. 
2.  The  standard  meter  is  a  bar  of  platinum  deposited  in  the 
archives  of  Paris. 

298.  The  Metric  System  employs  five  different  units  or 
denominations  and  seven  prefixes. 

The  units  are  the  me'ter,  li'ter,  gram,  ar,  and  ster.\ 

299.  The  names  of  the  higher  denominations  are  formed 
by  prefixing  to  the  unit  the  Greek  numerals,  dele'  a  10, 
hek'to  too,  kil'o  1000,  and  myr'ia  10000  ;  as,  dek'a-me'ter, 
10  meters;  hek'to-me'ter,  100  meters,  etc. 

The  names  of  the  lower  denominations,  or  divisions 
of  the  unit,  are  formed  by  prefixing  to  the  unit  the  Latin 
numerals,  dec'i  (des'-ee)  .  i,  cen'ti  (cent'-ee)  .01,  and  mil'li 
(mil'-lee)  .001;  as,  dec'i-me'ter,  T^  meter;  cen'ti-me'ter, 
meter  ;  mil'li-me'ter,  T  ^  0  meter. 


NOTE.  —  The  numeral  prefixes  are  the  key  to  the  whole  system  ; 
their  meaning  therefore  should  be  thoroughly  understood. 

296.  Upon  what  are  the  Metric  Weights  and  Measures  founded?  Why  eo 
called  ?  297.  What  is  the  meter  »  299.  How  are  the  names  of  the  lower  denom- 
inations formed  ?  The  higher  ? 

*  This  system  had  its  origin  in  France,  near  the  close  of  the  last  century.  Ita 
simplicity  and  comprehensiveness  have  secured  its  adoption  in  nearly  all  the 
countries  of  Europe  and  South  America.  Its  use  was  legalized  in  Great  Britain 
in  1864  ;  and  in  the  United  States,  by  Act  of  Congress,  1866. 

t  The  spelling,  pronunciation,  and  abbreviation  of  metric  terms  in  this  work, 
arf>  the  same  as  adopted  by  the  American  Metric  Bureau,  Boston,  aud  the, 
Mevrolo;;ical  Society,  New  York, 


208  METRIC     OB     DECIMAL 

LINEAR    MEASURE. 

300.  The  Unit  of  Length  is  the  METER,  which  is 
equal  to  39.37  inches.* 

The  denominations  are  the  mil'li-me'ter,  cen'ti-me'ter, 
dec'i-me'ter,  meter,  dek'a-me'ter,  hek'to-me'ter,  kil'o-me'ter, 
and  myr'ia-me'ter. 

TABLE. 

10  mtt'ttme'ters  (mm.)  make  i  cen'ti-me'ter  -    - .  cm. 

10  cen  ti  me'ters  "      i  dec'i-me'ter     -    -    dm. 

10  dec'i-me  ters  "      i  ME'TER   -    -    m. 

10  me'ters  "      i  dek'a-me  ter  -    -    Dm. 

10  dek'a-me'ters  "      I  hek'to-me'ter  -    -    Hm. 

10  hek'to-me  ters  "      i  kil'o-me'ter     •    -    Km. 

10  kil'o-me'ters  "      i  myr'ia-me'ter     -    Mm. 

NOTES. — i.  The  Accent  of  each  unit  and  prefix  is  on  the  first 
syllable,  and  remains  so  in  the  compound  words. 

To  abbreviate  the  compounds,  pronounce  only  the  prefix  and  the 
first  letterof  the  unit ;  as,  centim,  million,  centil,  decig,  hektog,  etc. 

•2.  The  meter,  like  our  yard,  is  used  in  measuring  cloths,  laces, 
moderate  distances,  etc. 

For  long  distances  the  kilometer  (3280  ft.  10  in.)  is  used  ;  and 
for  minute  measurements,  the  centimeter  or  millimeter. 

3.  Decimeters,  dekameters,  hektometers,  like  our  dimes  and  eagles, 
are  seldom  used. 

SQUARE    MEASURE. 

301.  The  Unit  for  measuring  surfaces  is  the  Square 
Meter,  which  is  equal  to  1550  sq.  in. 

The  denominations  are  the  sq.  cen'ti-me'ter,  sq.  dec'i- 
me' tjbr,  and  sq.  me' ter. 

100  sq.  cen'tim.  make  I  sq.  dec'i-me'ter  -    -    sq.  dm. 
100  sq.  dec'im.        "      i  Sq.  Me* ter      -    -    sg.  m. 

NOTE. — The  square  meter  is  used  in  measuring  floorings,  ceilings, 
etc.;  square  deci-mtters  and  eenti-meters,  for  minute  surfaces. 

300.  What  is  the  unit  of  Linear  Meaeare  ?  Its  denominations  ?  Recite  the 
Table.  301.  What  is  the  unit  for  measuring  surfaces  ?  Recite  the  table. 

*  Authorized  by  Act  of  Congress,  1866. 


WEIGHTS     AND     MEASURES.  209 

302.  The  Utt  it  for  measuring  land  is  the  Ar,  which 
is  equal  to  a  square  dekameter,  or  119.6  sq.  yards. 

The  only  subdivision  of  the  Ar  is  the  cen'tar ;  and 
the  only  multiple  is  the  liek'tar.  Thus, 

100  cent'ars  (ca.)  make  i  Ar    -    -    -    a. 
100  are  "       i  hek'tar  -    -    Ha. 

NOTES. — i.  The  term  ar  is  from  the  Latin  area,  a  surface. 

2.  In  Square  Measure,  it  takes  roo  units  of  a  lower  denomination 
to  make  one  in  the  next  higher  ;  hence  each  denomination  must  have 
two  places  of  figures.  In  this  respect  centars  correspond  to  cents. 

CUBIC    MEASURE. 

303.  The    Unit  for  measuring  solids  is   the   Cubic 
Meter,  which  is  equal  to  35.316  cu.  ft. 

TABLE. 

1000  cu.  mil'lim.  make  i  cu.  cen'ti  me'ter  -  cu.  cm. 
1000  cu.  cen'tim.  "  i  cu.  dec'i-me'ter  -  cu.  dm. 
1000  cu.  dec'im.  "  i  Cu.  Me'ter  -  cu.  m. 

NOTES. — i.  The  cubic  meter  is  used  in  measuring  embankments, 
excavations,  etc. ;  cubic  centimeters  and  millimeter tf.for  minute  bodies. 

2.  Since  it  takes  1000  units  of  a  lower  denomination  in  cubic 
measure  to  make  one  of  the  next  higher,  it  is  plain  that,  like  mills, 
each  denomination  requires  three  places  of  figures. 

304.  In  measuring  wood,  the  Ster,  which  is  equal  to 
a  cubic  meter,  is  sometimes  used. 

The  only  subdivision  of  the  ster  is  the  dec'i-ster  ;  and 
the  only  multiple,  the  dck'a-sfer.  Thus, 

10  dec'i-sters  make  i  Ster  -    -    -    s. 
10  sters  "      i  dek'a-ster    -    Ds. 

NOTES. — i.  The  term  ster  is  from  the  Greek  stereos,  a  solid. 
2.  In  France,  firewood  is  commonly  measured  in  a  cubical  box, 
whose  length,  breadth,  and  height  are  each  i  meter. 

3.  As  the  ster  is  applied  only  to  wood,  and  probably  will  never 
come  into  general  use,  its  divisions  and  multiples  may  be  omitted. 

Note.  How  many  places  of  figures  does  each  denomination  occupy?    Why? 

302.  What  is>  the  unit  for  measuring  l.inrt  ?     Its  divisions?      Its  multiples? 

303.  What  are  the  units  for  menf-urnis*  solids?    Recite  the  Table.     How  many 
places  does  each  denomination  in  cubic  measure  occupy  ?    Why  ? 


210  METRIC     OB     DECIMAL 


DRY    AND    LIQUID    MEASURE. 

305.  The  Unit  of  Dry  and  Liquid  Measure 

is  the  Liter  (lee'ter],  which  is  equal  to  a  cubic  decim,  or 
1.0567  liquid  quart  or  0.908  dry  quart. 

The  denominations  are  the  mil'li-li'ter,  cen'ti-li'ter, 
dec'i-li'ter,  liter,  dek'a-li'ter,  hek'to-li'ter,  kil'o-li'tei: 

10  mil'li-li'ters  (ml.)  make  i  cen'ti-li'ter  -  -  -  d. 

10  cen'ti-li'ters  "  i  dec'i-li'ter    -  -  -  dl. 

10  dec'i-li'ters  "  i  Li'ter  -    -  -  -  I. 

10  li'ters  "  i  dek'a-liter  -  -  -  Dl. 

10  dek'a-li'ters  "  i  hek'to-li'ter  -  -  HI. 

10  hek  to-li'ter  "  i  kil'o-li'ter  -  -  -  Kl. 

10  kil'o  liters  "  i  myr'ia-li'ter  -  -  Ml. 

NOTES. — i.  The  liter  is  used  in  measuring  milk,  wine,  etc.  For 
small  quantities,  the  centiliter  and  millUiter  are  employed  ;  and  for 
large  quantities,  the  dekaliter. 

2.  For  measuring  grain,  etc.,  the  hektoliter,  which  is  equal  to 
2.8375  bushels,  is  commonly  used. 

3.  The  term  liter  is  from  the  Greek  litra,  a  pound. 

WEIGHT. 

306.  The  Unit  of  Weight  is  the  Gram,  which  is 
equal  to  15.432  grains. 

The  denominations  are  the  mil'li-gram,  cen'ti-gram, 
dec'i-gram,  gram,  dek'a-gram,  hek'to-gram,  kil'o-gram, 
myr'ia-gram,  and  ton'neau  or  ton. 


10  milli  -grams  (ing. 

)  make  i  centi-gram    -    - 

eg. 

10  centi  grams 

"      i  dec'i-gram      -    - 

dg. 

10  dec  i-grams 

"      i  Grain 

ff- 

10  grams 

"      i  dek'a-gram    -    - 

Dg. 

10  dek'a-grams 

"      i  hekto-gram  -    - 

Hg. 

10  hek  to-grams 

"      i  Kil'o-grani   • 

Kg. 

10  kil  o-grams 

"       i  myr'ia-gram  -    - 

Mg. 

100  myr'ia-grams 

"       i  TONNEAU  or  Ton 

T. 

305.  What  is  the  unit  of  Dry  and  Liquid  Men^ure?    Its  denomlnatitmst    Ra- 
pnat  the  Table.    306.  What  is  the  unit  of  Weight?    Its  deuomiitutioimf 


WEIGHTS     AXD     MEASURES.  211 

REMARK. — As  the  quintal  (10  Mg.)  is  seldom  used,  and  millier 
means  the  same  as  tonneau  or  ton,  both  these  terms  may  be  dropped 
from  the  table.  The  metric  ton  is  equal  to  2204.6  Ibs. 

NOTES. — i.  This  Table  is  used  in  computing  the  weight  of  all 
objects,  from  the  minutest  atom  to  the  largest  heavenly  body. 

2.  The  grum  is  derived  from  the  Greek  gramma,  a  rule  or 
standard,  and  is  equal  to  a  cubic  centimeter  of  distilled  water  at  its 
greatest  density,  viz.,  at  the  temperature  of  4°  centigrade  thermome- 
ter, or  39.83°  Fahrenheit,  weighed  in  a  vacuum. 

3.  The  common  unit  for  weighing  groceries  and  coarse  articles 
is  the  kilogram,  which  is  equal  to  2.2046  pounds  Avoirdupois. 

4.  Kilogram  is  often  contracted  into  kilo,  and  tonneau  into  tan. 

307.  To  express  Metric  Denominations  decimally  in  terms 
of  a  given  Unit. 

1.  Write  7  Hm.  o  Dm.  9  m.  3  dm.  5  cm.  decimally  in 
terms  of  a  meter. 

ANALYSIS. — Metric    denominations    in-  OPERATION. 

crease  and  decrease  by  the  scale  of  10,  and  7°9'35  m->  A&8* 
correspond  to  the  orders  of  the  Arabic 

Notation.  We  therefore  write  the  given  meters  as  units,  the  deka- 
meters  as  tens,  and  the  hektometers  as  hundreds,  with  the  deci- 
meters and  centimeters  on  the  right  as  decimals.  Hence,  the 

RULE. —  Write  the  denominations  above  the  given  unit 
in  their  order,  on  the  left  of  a  decimal  point,  and  those 
below  the  unit,  on  the  right  as  decimals. 

NOTES — i.  If  any  intervening  denominations  are  omitted  in  the 
given  number,  their  places  must  be  supplied  by  ciphers. 

2.  As  each  denomination  in  square  measure  occupies  two  places  of 
figures,  in  writing  square  decimeters,  etc.,  as  decimals,  if  the  number 
is  less  than  10,  a  cipher  must  be  prefixed  to  the  figure  denoting  them. 

Thus,  17  sq.  in.,  and  2  sq.  dm.  =  17.02  sq.  meters.    (Art.  302,  ??.) 

3.  In  like  manner,  in  writing  cubic  decimeters,  etc.,  as  decimals, 
if  the  number  is  less  than  10,  two  ciphers  must  be  prefixed  to  it. 

Thus,  63  cu.  m.  and  5  cu  dm.  =  63.005  cu.  meters. 


Repeat  the  Table.  Note.  The  common  unit  for  weighing  groceries,  etc.? 
307.  How  write  metric  qrnnHti'js  In  terms  of  a  given  unit  5-  AV>V.  1C  any  clo- 
uomiitur.ion  is  omitted,  what  is  to  be  don«  ? 


212  METBIC     OR     DECIMAL 

308.  Metric  denominations  expressed  decimally  are  read 
like  numbers  in  the  Arabic  notation. 

Thus,  the  Answer  to  Ex.  i,  (709.35),  is  read,  "seven  hundred 
and  nine,  and  thirty-five  hundredths  meters,  or  seven  hundred 
nine  meters,  thirty-five,"  as  we  read  $709.35,  "seven  hundred 
nine  dollars,  thirty-five,"  omitting  the  name  cents. 

NOTE. — The  number  of  figures  following  the  name  of  the  unit, 
shows  the  denomination  of  the  last  decimal  figure. 

If  the  name  is  required,  it  is  better  to  use  abbreviations  of  the 
metric  terms,  as  centims,  millims,  etc.,  than  the  English  words 
hundredth*,  or  thousandths  of  a  meter,  etc.  The  former  are  not  only 
significant  of  the  kind  of  unit  and  the  value  of  the  decimal,  but 
are  understood  by  all  nations. 

Write  decimally,  and  read  the  following : 

1.  5  Mm.,  3  Km.,  7  Hm.,  o  Dm.,  9  m.,  8  dm.,  4  cm., 
5  mm.  in  terms  of  a  meter,  hektometer,  kilometer,  and 
decimeter. 

2.  4  Kg.,  5  Hg.,  o  Dg.,  5  g,  i  dc.,  o  eg.,  8  mg.  in  terms 
of  a  dekagram,  centigram,  hektogram,  and  kilogram. 

3.  Write  and  read  5  KL,  o  HI.,  6  Dl.,  3  1.,  o  dl.  8  cl. 
in  terms  of  a  hektoliter.          Ans.  50.6308  hektoliters. 

309.  To  reduce  higher  Metric  Denominations  to  lower. 
Ex.  i.  Eeduce  46.3275  kilometers  to  meters. 

ANALYSIS  — From  kilometers  to  meters  OPERATION. 

there    are    three   denominations,   and  it  4"-3275 

takes  10 of  a  lower  denomination  to  make  Ioo° 

a  unit  of  the  next  higher.     We  therefore     Ans.  46327.5000  m. 

multiply  by  1000,  or  remove  the  decimal 

point  three  places  to  the  right.     (Art.  181.)    Hence,  the 

RULE. — Multiply  the  given  denomination  by  10,  100, 
1000,  etc.,  as  the  case  way  require  ;  and  point  off  the  prod- 
uct as  in  multiplication  of  decimals.  (Art.  191.) 

NOTE. — It  should  be  remembered  that  in  the  Metric  System  each 
denomination  of  square  measure  requires  two  figures ;  and  each 
denomination  of  cubic  measure,  three  figures. 


WEIGHTS   AND   MEASUKES.  813 

e.   In  43.75  hectares,  how  many  square  meters? 

3.  Reduce  867  kilograms  to  grams. 

4.  Reduce  264.42  hectoliters  to  liters. 

5.  In  2561  ares,  how  many  square  meters? 

6.  In  865  2  cubic  meters,  how  many  cubic  decimeters  ? 

7.  Reduce  4256.25  kilograms  to  grams. 

310.  To  reduce  lower  Metric  Denominations  to  higher. 

8.  Reduce  84526.3  meters  to  kilometers. 

ANALYSIS. — Since  it  takes  10  linear  units 
to  make  one  of  the  next  higher  denomination,  'ERATION. 

it  follows  that  to  reduce  a  number  from  a     IOOO)84526-3  m. 
lower  to  the  next  higher  denomination,  it  84.5263  km. 

must  be  divided  by  10;  to  reduce  it  to  the 

next  higher  still,  it  must  be  again  divided  by  10,  and  so  on.  From 
meters  to  kilometers  there  are  three  denominations ;  we  therefore) 
divide  by  1000,  or  remove  the  decimal  point  three  places  to  the  left. 
The  answer  is  84.5263  km.  Hence,  the 

RULE. — Divide  the  given  denomination  by  10,  100,  1000, 
etc.,  as  the  case  may  require,  and  point  off  the  quotient  as 

in  division  of  decimals.     (Art.  194.) 

t 
9.  In  652254  square  meters,  how  many  hectares? 

10.  Reduce  87  meters  to  kilometers. 

n.  In  1482.35  grams,  how  many  kilograms? 

12.  In  39267.5  liters,  how  many  kiloliters? 


APPROXIMATE  VALUES. 

310  ft.  In  comparing  Metric  Weights  and  Measures  with  those 
now  in  use,  the  approximate  values  are  often  convenient.  Thus, 
when  no  great  accuracy  is  required,  we  may  consider 


i  decimeter  as 

4  in. 

i  cu.  meter  or  etere  as     j  cord. 

i  meter         " 

40  in. 

i  liter 

"       i  quart. 

5  meters        " 

i  rod. 

i  hectoliter 

"     2^  bushels. 

i  kilometer  " 

$  mile. 

i  gram 

"    15  i  grains. 

i  sq.  meter    " 

io£  sq.  feet. 

i  kilogram 

"      2  £  pounds. 

i  hectare       " 

2^  acres. 

i  metric  ton 

"  2200  pounds. 

309.  How  reduce  higher  metric  dtmoinmatioiia  to  lower  ?    310.  Lower  to  higher  1 


214  MKTEIC     OR     DECIMAL. 


APPLICATIONS    OF    METRIC  -  WEIGHTS      AND 
MEASURES. 

311.    To   add,  subtract,  multiply,  and   divide   Metric   Weights 

and  Measures. 

EULE. — Express  the  numbers  decimally,  and  proceed 
as  in  the  corresponding  operations  of  whole  numbers  and 
decimals. 

1.  What  is  the  sum  of  7358.356  meters,  8.614  hecto- 
meters, and  95  millimeters  ? 

SOLUTION. — 7358.356  m.  + 861.4 m.,  +.095  m.^Saig.Ssim.  Ans. 

2.  What  is  the  difference  between  8.5  kilograms  and 
976  grams?    ' 

SOLUTION. — 8.5  kilos— .976  kilos=7.524  kilos.  Ana. 

3.  How  much  silk  is  there  in  1 2\  pieces,  each  contain- 
ing 48.75  meters? 

SOLUTION. — 48.75  m.  x  12.5=609.375  m.  Ans. 

4.  How  many  cloaks,  each  containing  5.68  meters,  can 
be  made  from  426  meters  of  cloth  ? 

SOLUTION. — 426  rn.-r-5.68  m.=75  cloaks.  AJIS. 

312.  To  reduce  Metric  to  Common  Weights  and  Measures. 

i.  Eeduce  5.6  meters  to  inches. 

39.37  in. 
ANALYSIS. — Since  in  i  meter  there  are  39.37  inches,  c  6  m 

in  5.6  meters  there  are  5.6  times  39.37  in. ;  and 
39-37  x  5 .6=220.472  in.  Therefore,  in  5.6  m.  there 
are  220.472  in.  Hence,  the 

Ans.     220.472  in. 

EULE. — Multiply  tJie  value  of  th-e  principal  unit  of  the 
Table  by  the  given  metric  number. 

NOTE. — Before  multiplying,  the  metric  number  should  be  reduced 
to  the  same  denomination  as  the  principal  vnit.  whose  value  is 
taken  for  the  multiplicand. 


311.  How  are  Metric  Weights  and  Measures  added,  cnbtrsetod,  multiplied, 
and  divide-:?  ?    312.  How  reduce  metric  to  common  weights  and  measures  r 


WEIGHTS     AND     MEASURES.  2lo 

2.  In  45  kilos,  how  many  pounds?          Ans.  99.207  Ibs. 

3.  In  63  kilometers,  how  many  miles? 

4.  Eeduce  75  liters  to  gallons. 

5.  Reduce  56  dekaliters  to  bushels. 

6.  Reduce  120  grams  to  ounces. 

7.  Reduce  137.75  kilos  to  pounds. 

8.  In  36  ares,  how  many  square  rods  ? 

ANALYSIS. — In  I  are  there  are  1 19.6  sq.  yds. ;  hence  in  36  ares  there 
are  36  times  as  many.  Now  119.6x36=4305.6  sq.  yds.,  and  4305.0 
sq.  yds.-r- 30^=142.33  sq.  rods.  Ans. 

9.  In  60.25  hektars,  how  many  acres? 
10.  In  120  cu.  meters,  how  many  cu.  feet  ? 

313.  To  reduce  Common  to  Metric  Weights  and  Measures. 

n.  Reduce  213  feet  4  inches  to  meters. 

ANALYSIS— 213  ft.  4  in.=256o  in.    Now,  OPERAT10N 

in  39.37  in.  there  is  i  meter;  therefore,  in          .,7)2=;6o!oo  in. 
2560  in.  there  are  as  many  meters  as  39.37  is      ^"'^'  / — 3 — \ — 
contained  times  in  2560;  and  2560-7-39.37=  Ans.  65.02  +  00. 

65.02  +  meters.    Hence,  the 

RULE. — Divide  the  given  number  by  tlie  value  of  the 
principal  metric  unit  of  the  Table. 

NOTE. — Before  dividing,  the  given  number  should  bo  reduced  t« 
the  lowest  denomination  it  contains ;  then  to  the  denomination  IE 
which  the  value  of  the  principal  unit  is  expressed. 

1 2.  In  63  yds.  3  qrs.,  how  many  meters  ? 

13.  Reduce  13750  pounds  to  kilograms. 

14.  Reduce  250  quarts  to  liters. 

15.  Reduce  2056  bu.  3  pecks  to  kilohters. 

16.  In  3  cwt.  15  Ibs.  12  oz.,  how  many  kilos? 

17.  In  7176  sq.  yards,  how  many  sq.  meters? 

18.  In  40.471  acres,  how  many  hektars? 

19.  In  14506  cu.  feet,  how  many  cu.  meters? 

20.  In  36570  cu.  yards,  how  many  cu.  meters? 


313.  How  reduce  common  to  metric  weights  and  measures  t 


COMPOUND   ADDITION. 

314.  To  add  two  OP  more  Compound  Numbers. 

i.  What  is  the  sum  of  13  Ib.  7  oz.  3  pwt  18  gr. ;  9  oz.  5 
pwt.  6  gr. ;  and  2  Ib.  8  oz.  8  pwt.  3  gr.  ? 

ANALYSIS. — Writing  the  same  denomi-  OPERATION. 

nations  one  under  another,  and  beginning  lb-     oz-    P^     S1"- 

at  the  right,  3  gr.  +  6  gr.  +  i8  gr.  are  27 
gr.  The  next  higher  denomination  is 
pwt.,  and  since  24  gr.=i  pwt.,  27  gr. 

must  equal  I  pwt.  and  3  grains.  We  set  Ans.  1J  o  17  3 
the  3  gr.  under  the  column  added,  be- 
cause they  are  grains,  and  carry  the  i  pwt.  to  the  column  of  penny 
weights,  because  it  is  a  unit  of  that  column.  (Art.  30,  n.)  Next,  i 
pwt. +  8  pwt. +  5  pwt. +  3  pwt.  are  17  pwt.  As  17  pwt.  is  less  than 
an  ounce  or  a  unit  of  the  next  higher  denomination,  we  set  it  under 
the  column  added.  Again,  8  oz.  +  9  oz.  +  7  oz.  are  24  oz.  Now  as 
12  oz.=i  Ib.,  24  oz.  must  equal  2  Ib.  and  o  oz.  We  set  the  o  under 
the  ounces,  and  carry  the  2  Ib.  to  the  next  column.  Finally,  2  Ib. 
4-2  Ib.  + 13  Ib.  are  17  Ib.,  which  we  set  down  in  full.  Hence,  the 

RULE. — I.  Write  the  same  denominations  one  under 
another ;  and,  beginning  at  the  right,  add  each  column 
separately. 

II.  If  the  sum  of  a  column  is  less  than  a  unit  of  the 
next  higher  denomination,  write  it  under  the  column  added. 

If  equal  to  one  or  more  units  of  the  next  higher  denomi- 
nation, carry  these  units  to  that  denomination,  and  write 
the  excess  under  the  column  added,  as  in  Simple  Addition. 

NOTES. — i.  Addition,  Subtraction,  etc.,  of  Compound  Numbers  are 
the  same  in  principle  as  Simple  Numbers.  The  apparent  difference 
arises  from  their  scales  of  increase.  The  orders  of  the  latter  increase 
by  the  constant  scale  of  10 ;  the  denominations  of  the  former  by  a 
variable  scale.  In  both  we  carry  for  the  number  which  it  takes  of  a 
lower  order  or  denomination  to  make  a  unit  in  the  next  higher. 

2.  If  the  answer  contains  a  fraction  in  any  of  its  denominations, 
except  the  lowest,  it  should  be  reduced  to  wfwle  numbers  of  lower 
denominations,  and  be  added  to  those  of  the  same  name. 

314.  How  are  compound  numbers  added  ?  \ote.  The  difference  between  Com 
pound  and  Simple  Addition  ?  If  the  answer  contains  a  fraction  ? 


COMPOtTND     ADDITION.  217 

3  if  they  occur  in  the  given  numbers,  it  is  generally  best  to 
reduce  them  to  lowor  denominations  before  the  operation  is  com- 
menced, 

2.  What  is  the  sum  of  6  gal.  3  qt.  i  pt.  2  gi. ;  4  gal.  2  qt. 
o  pt  i  gi.,  and  7  gal.  i  qt.  i  pt.  3  gi.  ? 

Ans.  1 8  gal.  3  qt.  i  pt.  2  gi 

(3-)  (40  (5-) 


£ 

8. 

d. 

far. 

T. 

cwt. 

Ib. 

oz. 

bu. 

pk. 

qt. 

7 

16 

6 

I 

3 

6 

42 

4 

14 

2 

5 

i 

5 

8 

2 

7 

0 

26 

7 

21 

3 

6 

2 

7 

9 

3 

15 

17 

14 

8 

8 

i 

7 

6.  A  merchant  sold  19!  yd.  of  silk  to  one  lady;  i6|  yd. 
to  a  second,  20^  yd.  to  a  third,  and  17!  to  a  fourth :  how 
much  silk  did  he  sell  to  all  ? 

7.  If  a  dairy  woman  makes  315  Ib.  5  oz.  of  butter  in 
June,  275  Ib.  10  oz.  in  July,  238  Ib.  8  oz.  in  August,  and 
263  Ib.  14  oz.  in  September,  how  much  will  she  make  in 
4  months  ? 

8.  A  ship's  company  drew  from  a  cask  of  water  10  gals. 

2  qts.  i  pt.  one  day,  12  gal.  3  qt.  the  second,  15  gal.  i  qt. 
i  pt.  the  third,  and  16  gal.  3  qt.  the  fourth:  how  much 
water  was  drawn  from  the  cask  in  4  days  ? 

9.  A  farmer  sent  4  loads  of  wheat  to  market ;  the  ist 
contained  45  bu.  3  pk.  2  qt. ;  the  2d,  50  bu.  i  pk.  3  qt ; 
the  3d,  48  bu.  2  pk.  5  qt. ;  and  the  4th,  51  bu.  3  pk.  5  qt. : 
what  was  the  amount  sent  ? 

10.  How  much  wood  in   5  loads  which   contain   re- 
spectively i  c.  28^  cu.  ft.;  i  c.  47!  cu.  ft;  i  c.  n  cu.  ft. 

i  c.  no  cu.  ft;  and  i  c.  n-J  cu.  ft.  ? 

1 1.  A  miller  bought  3  loads  of  wheat,  containing  28  bu. 

3  pks. ;  36  bu.  i  pk.  7  qts. ;  33  bu.  2  pks.  3  qts.  respectively: 
how  many  bushels  did  he  buy  ? 

If  any  of  the  given  numbers  contain  a  fraction,  how  proceed?    315.  How  add 
denominate  fractions  * 

10 


218  COMPOUND     ADDITION. 

12.  What  is  the  sum  of  13  m.  65  r.  3  yd.  i  ft.  ;  12  m. 
40  r.  2  yd.  i  ft.  ;  and  21  m.  19  r.  i  yd.  2  ft.  ? 

SOLUTION.  —  In  dividing  the  yards  m.      r.  yd.    ft. 

by   s£   (the   number   required   to  *3      °5  3      I 

ruake  a  rod),  the  remainder  is  i£  12      40  2      I 

yard.    But  i£  yd.  equals  i  ft.  6  in.  ;  21      19  i  _  2 

we  therefore  add  i  ft.  6  in.  to  the  46    125  li  i 

feet  and  inches  in  the  result.    The  I  yd.   =  o      I      6  in. 

Ana.  5s  46  m.  I25  r.  i  yd.  2  ft.  6  in.     ^s.  46   125  i      2     6 


13.  What  is  the  sum  of  8  w.  2^«d.  7  h.  40  m.  ;  4  w.  5^  d. 
o  h.  15  m.  ;  and  10  w.  o  d.  16  h.  3  m.  ? 

14.  What  is  the  sum  of  15  A.  no  sq.  r.  30  sq.  yds.; 
6  A.  45  sq.  r.  16  sq.  yds  ;  42  A.  10  sq.  r.  25  sq.  yds.  ? 

315.  To  add  two  OP  more  Denominate  Fractions. 

15.  What  is  the  sum  of  f  bu.  £  pk.  and  f  quart  ? 
ANALYSIS  -Reducing  the   de-        4  bu>          pk.   x  qi   ,,     t> 

nominate  fractions  to  whole  num-        f     ^  _  ": 

bers,     then     adding    them,    the        3  nf    _  o  iJL 

amount  is  3  pk.  6  qt.  Oj^  pt., 


CUU.I/ U.J-1.  V     J.O     3     I^JX-      V    V^W*      VYO     ,fu'>    -rfC-l.  **O.  J  j_ 

(Art.  292.)  °Ttr 

1 6.  What  is  the  sum  of  £.125,  .655.  and  .75d.  ? 

ANALYSIS. — Reducing  the  denominate  £.125  —  2S.  6d.  o  far 
decimals  to  whole  numbers  of  lower  .658.  =o  7  3.2 

denominations,  then  adding,  the  result  .75d.  — o     o      3 

is  33.  2d.  2.2  far.     Hence,  the 

jfift/G.       3       ~        &.£ 

KULE. — Reduce  the  denominate  fractions,  whether  com- 
mon or  decimal,  to  whole  numbers  of  lower  denominations : 
then  add  them  like  other  compound  numbers. 

17.  Add  .75  T.  .5  cwt.  .25  Ib.    19.  Add  f  Ib.  ^X0  oz.  -^  pwt. 

18.  Add  .25  bu.  \  of  3  pk.        20.  Add  £.15,  .55.  .8d. 

21.  Bought  4  town   lots,  containing  \  acre;  f  acre; 
i2i£  sq.  rods;  and  150.5  sq.  rods  respectively:  how  much 
land  was  there  in  the  4  lots  ? 

22.  Bought  3  loads  of  wood ;  one  containing  £  cord, 
another  75.3  cu.  feet;  the  other  ij  cord:  how  much  did 
they  all  contain  ? 


COMPOUND   SUBTKACTIOK'. 

316.  To  find  the  Difference  between  two  Compound  Numbers. 

1.  From  5  mi.  3  fur.  n  r.  4  ft.  take  2  mi.  i  fur.  4  r.  9  ft. 
ANALYSIS. — We  write  the  less  number  m.    fur.     r.        ft. 

under  the  greater,  feet  under  feet,  etc.  S311        4 

Beginning  at  the  right,  9  ft.  cannot  be  2      I        4       9 

taken  from  4  feet ;  we  therefore  borrow  i  Ans.  3  2  6  I  li 
rod,  a  unit  of  the  next  higher  denomina- 
tion. Now  i  rod,  or  i6V  ft.,  added  to  4  ft.  are  20^  ft.,  and  9  ft.  from 
zo\  ft.  leave  n£  ft.  Set  the  remainder  under  the  term  subtracted, 
and  carry  i  to  the  next  term  in  the  lower  number,  i  rod  and  4  r.  are 
5  r.,  and  5  r.  from  n  r.  leave  6  rods,  i  furlong  from  3  fur.  leaves 
2  fur. ;  and  2  m.  from  5  m.  leave  3  m.  Therefore,  etc.  Hence,  the 

RULE. — I.  Write  the  less  number  under  the  greater, 
placing  the  same  denominations  one  under  another. 

II.  Beginning  at  the  right,  subtract  each  term  in  the 
lotver  number  from  that  above  it,  and  set  the  remainder 
under  the  term  subtracted. 

III.  If  any  term  in  the  lower  number  is  larger  than  that 
above  it,  borrow  a  unit  of  the  next  higher  denomination  and 
add  it  to  the  upper  term  ;  then  subtract,  and  carry  i  to  the 
next  term  in  the  lower  number,  as  in  Simple  Subtraction. 

NOTES. — i.  Sorrowing  in  Compound  Subtraction  is  based  on  the 
same  principle  as  in  Simple  numbers.  That  is,  we  add  as 
many  units  to  the  term  in  the  upper  number  as  are  required  of 
that  denomination  to  make  a  unit  of  the  next  higher. 

2.  If  the  answer  contains  a  fraction  in  any  of  its  denominations 
except  the  lowest,  it  should  be  reduced  to  lower  denominations,  and 
be  united  to  those  of  the  same  name,  as  in  Compound  Addition. 

2.  From  i  mile,  take  240  rods  3  yd.  2  ft. 

3.  From  14  Ib.  5  oz.  3  pwt.  16  gr.,  take  7  Ib.  6  pwt.  7  gr. 

4.  From  12  T.  9  cwt.  41  Ib.,  take  5  T.  i  cwt.  15  Ib. 

5.  Take  20  gal.  3  qt.  i  pt.  2  gi.  from  29  gal.  i  qt.  i  pt. 

316.  How  subtract  compound  numbers?    Note.  What  is  said  of  borrowing 
If  the  answer  contains  a  fraction,  how  proceed  ? 


220  COMPOUND     STTBTBACIION. 

6.  From  250  A.  45  sq.  r.  150  sq.  ft.,  take  91  A.  32  sq.  t 
200  sq.  ft. 

7.  A  miller  has  two  rectangular  bins,  one  containing 
324  cu.  ft.  no  cu.  in.;  the  other,  277  cu.  ft.  149  cu.  in.: 
#hat  is  the  difference  in  their  capacity  ? 

8.  A  man  having  320  A.  50  sq.  r.  of  land,  gave  his 
eldest  son  175  A.  29  sq.  r.,  and  the  balance  to  his  youngest 
son :  what  was  the  portion  of  the  younger  ? 

9.  A  merchant  bought  two  pieces  of  silk ;  the  longer 
contained  57  yd.  3  qr. ;  the  difference  between  them  was 
i if  yards  :  what  was  the  length  of  the  shorter? 

10.  What  is  the  difference  between   10  in.  2  fur.  27  r. 
i  ft  7  in.  and  6  m.  4  fur.  28  r.  i  yd.  ? 

317.  To  find  the   Difference   between   two    Denominate  Frac- 
tions. 

1.  From  £  of  a  yard,  take  -fj  of  a  foot. 

ANALYSIS. — Reducing  the  denominate  £  yd.  =  2  f t.  6  in. 

fractions  to  whole    numbers,   we    have  li  ft.    =  o  1 1  in. 

|  yard  =  2  ft.  6  in.  and  -^  foot  =   n  in.  ~~j~           i  ft.  TllT. 
Ana.  i  ft.  7  in.    Hence,  the 

RULE. — Reduce  the  denominate  fractions,  whether  com- 
mon or  decimal,  to  whole  numbers  of  lower  denominations  ; 
then  proceed  according  to  the  rule  above. 

2.  From  £.525  take  .75  of  a  shilling. 

3.  From  I  bu.  take  -f  of  a  peck. 

4.  From  |  of  a  gal.,  take  i  f  of  a  pint. 

5.  A  goldsmith  having  a  bur  of  gold  weighing  1.25  Ibs., 
cnt  off  sufficient  to  make  6  rings,  each  weighing  .15 
ounce ;  how  much  was  left 9 

6.  If  a  man  has  .875   acre  of  land,  and  sells  off  two 
building  lots,  each  containing   12.8  sq.  rods,  how  much 
land  will  be  left  ? 

7.  From  f  of  f  cwt.  of  sugar,  take  31.25  Ibs. 

317.  How  subtract  denominate  fractions  ? 


COMPOUND     SUBTRACTION.  221 

318.  To   find   the   Difference   of  Time   between   two   Dates, 

in  years,  months,  and  days. 

i.  What  is  the  difference  of  time  between  July  4th, 
1776,  and  April  3d,  1870? 

ANALYSIS. — We  place  the  earlier  date  tin-  y.  m.  d. 

der  the  later,  with  the  year  on  the  left,  the  1870  4  3 

number  of  the  month  next,  and  the  day  of  I77&  7  4 

the  month  on  the  right.     Since  we  cannot  Ans.     93  8  29 
take  4  from  3,  we  borrow  30  days,  and  4  from 

33  leaves  29.     Carrying  i  to  the  next  denomination,  we  proceed  as 
above.     (Art.  264,  n.)     Hence,  the 

KULE. — Set  the  earlier  date  under  the  later,  the  years  on 
the  left,  the  month  next,  and  the  day  on  the  right,  and  pro- 
ceed as  in  subtracting  other  compound  numbers. 

NOTE. — Centuries  are  numbered  in  dates,  from  the  beginning  of 
the  Christian  era ;  months,  from  the  beginning  of  the  year ;  and 
days,  from  the  beginning  of  the  month. 

3.  Washington  was  born  Feb.  22d,  1732,  and  died  Dec. 
1 4th,  1799:  at  what  age  did  he  die? 

319.  To  find   the   difference    between   two   Dates   in   Days, 

the  time  being  less  than  a  year. 

i.  A  note  dated  Oct.  2oth,  1869,  was  paid  Feb.  icth, 
1870 :  how  many  days  did  it  run  ? 

ANALYSIS. — In  Oct.  it    ran    31—20=11  Oct.  31  — 20— n  d. 

days,  omitting  the  day  of  the  date ;  in  Nov.,  Nov.                =  30  d. 

30  days;  in  Dec.,  31  ;  in  Jan.,  31;  and  in  Dec.                — 31  d. 

Feb.,    10,   counting   the  day   it   was  paid.  Jan.                —  31  d. 

Now  ii  +  30  +31  +  31  +  10=113  days,  the  peb.                =10  d. 

number  required.      Hence,  the  *                       i  it  d 

RULE. — Set  doion  the  number  of  days  in  each  month  and 
part  of  a  month  between  the  two  dates,  and  the  sum  will  be 
the  number  of  days  required. 

NOTE. — The  day  on  which  a  note  or  draft  is  dated,  and  that  on 
which  it  becomes  due,  must  not  both  be  reckoned.  It  is  customary 
to  omit  the  former,  and  count  the  latter. 


318.  How  find  the  difference  between  two  dates  in  years?    319.  In  days! 


222  COMPOUND     SUBTRACT  10  If. 

2.  What  is  the  number  of  days  between  Nov.  loth,  1869, 
and  March  3d,  1870? 

3.  A  person  started  on  a  journey  Aug.  ipth,  1869,  and 
returned  Nov.  ist,  1869:  how  long  was  he  absent? 

4.  A  note  dated  Jan.  3ist,  1870,  was  paid  June  3oth, 
1870:  how  many  days  did  it  run  ? 

5.  How  many  days  from  May  2ist,  1868,  to  Dec.  3ist, 
following  ? 

6.  The  building  of  a  school-house  was  commenced  April 
i  st,  and  completed  on  the   loth  of  July  following :  how 
long  was  it  in  building  ? 

320.  To  find  the  difference  of  Latitude  OP  Longitude. 

DBF.  i. — Latitude  is  distance  from  the  Equator.  It  is  reckoned 
in  degrees,  minutes,  etc.,  and  is  called  North  or  South  latitude,  ac- 
cording as  it  is  north  or  south  of  the  equator. 

2.  LoHf/ifnde  is  the  distance  on  the  equator  between  a  con <•»//- 
tionnl  or  fixed  meridian  and  the  meridian  of  a  given  place.  It  is 
reckoned  in  degrees,  minutes,  etc.,  and  is  called  East  and  West  longi- 
tude, according  as  the  place  is  east  or  west  of  the  fixed  meridian,  xmtil 
i8oD  or  half  the  circumference  of  the  earth  is  reached. 

7.  The  latitude  of  New  York  is  40°  42'  43"  N. ;  that  of 
New  Orleans,  29°  57'  30"  N.:  how  much  further  north  is 
New  York  than  New  Orleans  ?  * 

ANALYSIS. — Placing  the  lower  latitude  N.  Y.  40°  42'  43" 
under  the  higher,  and  subtracting  as  in  the  Jf.  Q.  29°  57'  30" 
preceding  rule,  the  Ans.  is  10°  45'  13".  Ans.  10°  Zq'  17" 

8.  The  latitude  of  Cape  Horn  is  55°  59'  S.,  and  that  of 
Cape  Cod  42°  i'  57.1"  N. :  what  is  the  difference  of  lati- 
tude between  them  ? 

ANALYSIS. — As  one  of  these  capes  is  north  of  the  equator  and  the 
other  smith,  it  is  plain  that  the  difference  of  their  latitude  is  the  sum 
of  the  two  distances  from  the  equator.  We  therefore  add  the  two 
latitudes  together,  and  the  result  is  98°  o'  57  i'. 

320.  Def.  What  is  latitude  ?  Longitude  ?  How  find  the  difference  of  latitude 
or  longitude  befween  two  places  ? 

*  The  latitude  and  longitude  of  the  places  in  the  United  States  here  given  are 
taken  from  the  American  Almanac,  1854 ;  those  of  places  in  foreign  lands,  montly 
from  Bo'-vditch'rf  Navigator. 


COMPOUND     SUBTRACTION.  223 

9.  The  longitude  of  Paris  is  2°  20'  East  from  Green- 
wich; that  of  Duolin  is  6°  20'  30''  West:  what  is  the  dif- 
ference in  their  longitude  ? 

ANALYSIS. — As  the  longitude  of  one  of  these  places  2°  20'  oo" 

is  East  from  Greenwich  the  standard  meridian,  and  6°  20'  30" 

that  of  the  other  West,  the  difference  of  their  longi-     7T5 ~, r, 

tude  must  be  the  sum  of  the  two  distances  from  the 
standard  meridian.    Ana.  8"  40'  30".    Hence,  the 

RULE. — I.  If  both  places  are  the  same  side  of  the  equator, 
cr  the  standard  meridian,  subtract  the  less  latitude  or 
longitude  from  the  greater. 

II.  If  the  places  are  on  different  sides  of  the  equator,  or 
the  standard  meridian,  add  the  two  latitiides  or  longitudes 
together,  and  the  sum  will  be  the  answer. 

10.  The  longitude  of  Berlin  is  13°  24'  E.  ;*   that  of 
New  Haven,  Ct.,  72°  55'  24"  "W. :  what  is  the  difference? 

11.  The  longitude  of  Cambridge,  Mass.,  is  71°  7'  22"; 
that  of  Charlottesville,  Va.,  78°  31'  29":    what  is  the 
difference  ? 

12.  The  latitude  of  St.  Petersburg  is  59°  56'  north,  that 
of  Rome  41°  54'  north :  what  is  their  difference  ? 

13.  The  latitude  of  Albany  is  4?°  39'  3",  that  of  Rich- 
mond 37°  32'  17":  what  is  the  difference? 

14.  The  latitude  of  St.  Augustine,  Flor.,  is  29°  48'  30" ; 
that  of  St.  Paul,  Min.,  is  44°  52'  46":  what  is  the  dif- 
ference ? 

15.  The  longitude  of  Edinburgh  is  3°  12'  W.;  that  of 
Vienna  16°  23'  E.f :  required  their  difference? 

1 6.  The  latitude  of  Valparaiso  is  33°  2'  S. ;  that  of  Ha- 
vana is  23°  9'  N.:  what  is  the  difference? 

17.  The  latitude  of  Cape  of  Good  Hope  is  34°  22'  S. ; 
that  of  Gibraltar,  36°  7'  N. :  what  is  their  difference  ? 

18.  The  longitude  of  St.  Louis  is  90°  15'  16";  that  of 
Charleston,  S.  C.,  79°  55'  38":  what  is  the  difference? 

*  Encyc.  Brit.  *  Encyc.  Amer. 


COMPOUND   MULTIPLICATION. 

321.  To  multiply  Compound  Numbers, 
i.  A  farmer  raised  6  acres  of  wheat,  which  yielded  15  bn. 
3  pk.  i  qt.  per  acre:  how  much  wheat  had  he  ? 

ANALYSIS. — 6  acres  will  produce  6  times  as  OPERATION. 

much  as  i  acre.     Beginning  at  the  right,  6  times  bu-    pk-  1*- 

1  qt.  are  6  qts.     As  6  quarts  are  less  than  a 
peck,  the  next  higher  denomination,  we  set  the 

6  under  the  term  multiplied.     6  times  3  pk.  are     Ans.  94      2      6 
18  pk.     Since  4  pk.=i  bu.,  18  pecks=4  bu.  and 

2  pk.  over.     Setting  the  remainder  2  under  the  term  multiplied,  and 
carrying  the  4  bu.  to  the  next  product,  we  have  6  times  15  bu.— go 
bu.,  and  4  bu.  make  94  bu.     Therefore,  etc.     Hence,  the 

RULE. — I.  Write  the  multiplier  under  the  lowest  denomi- 
nation of  the  multiplicand,  and,  beginning  at  the  right, 
multi2ily  each  term  in  succession. 

II.  If  the  product  of  any  term  is  less  than  a  unit  of  the 
next  higher  denomination,  set  it  under  the  term  multiplied. 

III.  If  equal  to  one  or  more  units  of  the  next  higher  de- 
nomination, carry  these  units  to  that  denomination,  and 
write  the  excess  under  the  term  multiplied. 

NOTES. — i.  If  the  multiplier  is  a  composite  number,  multiply  by 
one  of  the  factors,  then  this  partial  product  by  another,  and  so  on. 

2.  If  a  fraction  occurs  in  the  product  of  any  denomination  except 
tlnj  lowest,  it  should  be  reduced  to  lower  denominations,  and  be 
united  to  those  of  the  same  name  as  in  Compound  Addition.  (Art.  314.) 

(2.)  (3-) 

Mult.    12  T.  7  cwt.  1 6  Ib.  £21,  138.  8|d. 

By         8  7_ 

<j..  What  is  the  weight  of  10  silver  spoons,  each  weigh- 
ing 3  oz.  7  pwt.  13  gr.  ? 

321.  How  are  compound  numbers'  multiplied?  Nrce.  If  the  multiplier  is  a 
composite  number,  how  proceed  1  When  a  fraction  occurs  in  ai»y7  denomination 
except  tire  last,  how  ? 


COMPOUND  MULTIPLICATION.              225 

(50  (6.) 
Mult.  9  oz.  13  pwt.  7  gr.  by  18.  Mult.  10  r.  i  yd.  i  ft.  by  7. 

9  oz.  13  pwt.  7  gr.  10  r.  i  yd.  i  ft.  o  in. 

3  7 

2lb.  4    19    21  71   3^   i 

6  i)T«l.=        i   6 


14  Ib.  5  oz.  19  pwt.  6  gr.  Ans.   Ans.  71  r.  3  yd.  2  ft.  6  in. 

7.  If  a  family  use  27  gal.  2  qt.  i  pt.  of  milk  in  a  month, 
how  much  will  they  use  in  a  year  ? 

8.  If  a  man  chops  2  cords  67  cu.  feet  of  wood  per  day, 
how  much  will  he  chop  in  9  days  ? 

9.  What  cost  27  yards  of  silk,  at  173.  7^d.  sterling  per 
yard? 

10.  If  a  railroad  train  goes  at  the  rate  of  23  m.  3  fur. 
21  r.  an  hour,  how  far  will  it  go  in  24  hours? 

n.  How  much  corn  will  63  acres  of  land  produce,  at 
30  bu.  3  pk.  per  acre  ? 

1 2.  How  many  cords  of  wood  in  1 7  loads,  each  contain- 
ing i  cord  41  cu.  ft.? 

13.  How  much  hay  in  12  stacks  of  5  tons,  237  Ibs.  each? 

14.  How  much  paper  is  required  to  print  20  editions  of 
a  book,  requiring  65  reams,  7  quires,  and  10  sheets  each  ? 

15.  If  a  meteor  moves  through  5°  23' 15 ''in  a  second, 
how  far  will  it  move  in  30  seconds  ? 

1 6.  If  the  daily  session  of  a  school  is  5  h.  45  min.,  how 
many  school  hours  in  a  term  of  15  weeks  of  5  days  each  ? 

17.  A  man  has  u  village  lots,  each  containing  12  sq.  r. 
4  sq.  yd.  6  sq.  ft. :  how  much  do  all  contain  ? 

1 8.  If  i  load  of  coal  weighs  i  T.  48^  Ib.,  what  will  72 
loads  weigh  ? 

19.  How  many  bushels  of  corn  in  12  bins,  each  con- 
taining 130  bu.  3  pk.  and  7  qfc.  ? 

20.  A  grocer  bought  35  casks  of  molasses,  each  contain' 
ing  55  gal.  2  qt.  i  pt. :  how  much  did  they  all  contain? 


COMPOUKD  DIVISION. 

322.  Division  of  Compound  Numbers,  like  Simple 
Division,  embraces  two  classes  of  problems  : 

First.  —  Those  in  which  the  dividend  is  a  compound  number,  and 
the  divisor  is  an  abstract  number. 

Second.  —  Those  in  which  both  the  divisor  and  dividend  are  com- 
pound numbers.  In  the  former  the  quotient  is  a  compound  number. 
In  the  latter,  it  is  times,  or  an  abstract  number.  (Art.  64.) 

323.  To  divide  one  Compound  Number  by  another,  or  by  an 
Abstract  number. 

Ex.  i.  A  dairy-woman  packed  94  Ib.  2  oz.  of  butter  in 
6  equal  jars  :  how  much  did  each  jar  contain  ? 

ANALYSIS.  —  The  number  of  parts  is  given,  OPERATION. 

to  find  the  value  of  each  part.     Since  6  jars  6)94  Ib.  2  OZ. 

contain  94  Ibs.  2  oz.,  i  jar  must  contain  ^     Ans.     je  lb.il  OZ. 
of  94  Ibs.  2  oz.     Now  %  of  94  Ibs.  is  15  Ibs. 

and  4  Ibs.  remainder.     Reducing  the  remainder  to  oz.,  and  adding 
the  2  oz.,  we  have  66  oz.     Now  ^  of  66  oz.  is  n  oz.  (Art.  63,  6.) 

2.  A  dairy-woman  packed  94  Ibs.  2  oz.  of  butter  in  jars 
of  15  Ibs.  ii  oz.  each  :  how  many  jars  did  she  have? 

ANALYSIS.  —  Here  the  size  of  each  94  Ib.     2  oz.  =  1506  oz. 

part  is  given,  to  find  the  number  of  i  e  Ib.  1  1  oz.  =     251  oz. 

parts  in  94  Ib.  2  oz.     Reduce  both  OZ.)l«5o6  OZ. 

numbers  to  oz.,  and  divide  as  in  sim-  — 


.,  . 

pie  numbers.  (Art.  63,  a.)  Hence,  the  -*** 

fi,ULE.  —  I.  When  the  divisor  is  an  abstract  number, 
Beginning  at  the  left,  divide  each  denomination  in  suc- 

cession, and  set  the  quotient  tinder  the  term  divided. 

If  there  is  a  remainder,  reduce  it  to  the  next  lower  de- 

nomination, and,  adding  it  to  the  given  units  of  this  de- 

nomination, divide  as  before. 
II.  When  the  divisor  is  a  compound  number, 
Reduce  the  divisor  and  dividend  to  the  lowest  denomina- 

tion contained  in  either,  and  divide  as  in  simple  numbers. 

322.  How  many  classes  of  examples  docs  division  of  compound  numbers  em 
brace?    The  first?    The  second?    323.  What  is  the  rule  ? 


TIME     AND     LONGITUDE.  227 

NOTE. — If  the  divisor  is  a  composite  number,  we  may  divide  by  it* 
factors,  as  in  simple  numbers.  (Art.  77.) 

3.  Divide  29  fur.  19  r.  2  yd.  i  ft.  by  7. 

4.  Divide  54  gal.  3  qt.  i  pt.  3  gi.  by  8. 

5.  A  miller  stored  450  bu.  3  pks.  of  grain  in  18  equal 
bins :  how  much  did  he  put  in  a  bin  ? 

6.  A  farm  of  360  A.  42  sq.  r.  is  divided  into  23  equal 
pastures:  how  much  land  does  each  contain? 

7.  How  many  spoons,  each  weighing  2  oz.  10  pwt.,  can 
be  made  out  of  5  Ib.  6  oz.  of  silver  ? 

8.  How  many  iron  rails,  18  ft.  long,  are  required  for  a 
railroad  track  15  miles  in  length? 

9.  How  many  times  does  a  car-wheel  15  ft.  6  in.  in  cir- 
cumference turn  round  in  3  m.  25  r.  10  ft.  ? 

10.  How  many  books,  at  48.  6£d.  apiece,  can  you  buy  for 
£2,  143.  3d.? 

ii  If  6  men  mow  86  A.  64  sq.  rods  in  6  days,  how 
much  will  i  man  mow  in  i  day? 

12.  A  farmer  gathered  150  bu.  3  pk.  of  apples  from  24 
trees :  what  was  the  average  per  tree  ? 

Y 

COMPARISON    OF   TIME   AND   LONGITUDE. 

324.  The  Earth  makes  a  revolution  on  its  axis  once  in  24 
hours ;  hence  -*1,-  part  of  its  circumference  must  pass  under  the  sun 
in  i  hour.  But  the  circumference  of  every  circle  is  divided  into 
360° ,  and  vl4-  of  360°  is  15°.  It  follows,  therefore,  that  15°  of  longi- 
tude make  a  difference  of  i  hour  in  time. 

Again,  since  15"  of  longitude  make  a  difference  of  i  hour  in  time, 
15  of  longitude  (,/u  of  15")  will  make  a  difference  of  i  minute  (fa  of 
an  hour)  in  time. 

In  like  manner,  15"  of  longitude  (-^  of  15'),  will  make  a  difference 
of  i  second  iu  time.  Hence,  the  following 

TABLE. 

15°  of  longitude  are  equivalent  to  i  hour  of  time. 

J5'  "  i  minute       " 

15"  *  *          "  i  second       " 


228  TIME     AND     LONGITUDE. 

CASE    I. 

325.  To  find  the  Difference  of  Time  between  two  places, 
the  difference  of  Longitude  being  given. 

Ex.  i.  The  difference  of  longitude  between  New  York 
and  London  is  73°  54'  3":  what  is  the  difference  of  time? 

ANALYSIS. — Since  15°  of  Ion.  are  equiv-  OPERATION. 

alent  to  i  hour  of  time,  the  difference  of  15  j  73°  54'  3" 

time  must  be  -,V  part  as  many  hours,  min-      . 

A,       '  Ans.  4h.  55  m.  36.23. 

utes,  and  seconds  as  there  are  degrees, 

etc.,  in  the  dif.  of  Ion. ;  and  73°  54'  3"-j-i5=4  h.  55'  36.2 "     Hence,  the 

RULE. — Divide  the  difference  of  longitude  by  15,  and  the 
degrees,  minutes,  and  seconds  of  the  quotient  will  be  the 
difference  of  time  in  hours,  minutes,  and  seconds.  (Art.  3  23.) 

2.  The  difference  of  longitude  between  Savannah,  Ga., 
and  Portland,  Me.,  is  10°  53'  2":  what  is  the  difference  in 
time? 

i°  3'  30"  W.,  that  of 
,  noon  in  Boston  what 
is  the  time  at  Detroit  ? 

4.  The  longitude  of  Philadelphia  is  75°  9'  54",  that  of 
Cincinnati  84°  27':  when  it  is  noon  at  Cincinnati  what  is 
the  time  at  Philadelphia  ? 

5.  The  Ion.  of  Louisville,  Ky.,  is  85°  30',  that  of  Bur- 
lington, Vt.,  73°  10':  what  is  the  difference  in  time? 

6.  When  it  is  noon  at  Washington,  what  is  the  time  of 
(by  at  all  places  22°  30'  east  of  it?    What,  at  all  places 
22°  30'  west  of  it? 

7.  How  much  earlier  does  the  sun  rise  in  New  York, 
Ion.  74°  3",  than  at  Chicago,  Ion.  87°  35'? 

8.  How  much  later  does  the  sun  set  at  St.  Louis,  whoso 
longitude  is  90°   15'   16"  W.,  than  at  Nashville,  Tenn., 
whose  longitude  is  86°  49'  3"  ? 

325.  How  find  the  difference  of  time  between  two  places,  the  difference  ol 
longitude  being  given? 


3.  The  longitude  of  Boston  is  7 
Detroit  is  83°  2'  30"  W. :  when  it  is 


AND     LONGITUDB. 


CASE    II. 

326.   To  find  the  Difference  of  Longitude  between  two 
places,  the  difference  of  Time  being  given. 

9.  A  whaleman  wrecked  on  an  Island  in  the  Pacific, 
found  that  the  difference  of  time  between  the  Island  and 
San  Francisco  was  2  hr.  27  min.  54!  sec.  :  how  many  de- 
grees of  longitude  was  he  from  San  Francisco  ? 

ANALYSIS.  —  Since  15°  of  Ion.  are  equira  OPERATION. 

lent  to  i  hour  of  time,  15'  of  Ion.  to  i  min.  2\\.  27  m.  54.6  860. 

of  time,  and  15"  to  i  sec.  of  time,  there  must  t  r 

be  15  times  as  many  degrees,  minutes,  and  - 

seconds  in   the  difference   of  longitude   as  •»***•  3"    5°    39 
there  are  hours,  minutes,  and  seconds  in  the 
difference  of  time  ;  and  2  h.  27  m.  54.6  s.  x  15=36°  58'  39".   Hence,  the 

RULE.  —  Multiply  the  difference  of  time  fyi  i$>*nd  the 
hours,  minutes,  and  seconds  of  the  product  will  be  the 
difference  of  longitude  in  degrees,  minutes,  and  seconds. 

10.  The  difference  of  time  between  Richmond,  Va.,  ana 
Newport,  R,  I.,  is  24  min.  36  sec.  :  what  is  the  difference 
of  longitude  ? 

1  1.  The  difference  of  time  between  Mobile  and  Galveston 
is  27  min.  £  sec.:  what  is  the  difference  of  longitude? 

12.  The  difference  of  time  between  Washington  and 
San  Francisco  is  3  hr.  i  min.  39  sec.  :  what  is  the  dif- 
ference in  longitude  ? 

13.  The  distance  from  Albany,  N.  Y.,  to  Milwaukee,  is 
nearly  625  miles,  and  a  degree  of  longitude  at  these  places 
is  about  44  miles  :  how  much  faster  is  the  time  at  Albany 
than  at  Milwaukee  ? 

14.  The  distance  from  Trenton,  N.  J.,  to  Columbus,  0.. 
is  nearly  400  miles,  and  a  degree  of  longitude  at  these 
places  is  about  46  miles:  when  it  is  noon  at  Columbus 
what  is  the  time  at  Trenton  ? 

326.  How  find  the  different  of  longitude,  when  the  difference  of  time  U  given* 


PERCENTAGE. 

327.  Per  Cent  and  Rat  e  Per  Cent  denote  hun- 
dredths.    Thus,  i  per  cent  of  a  number  is  y^  part  of  that 
number  ;  3  per  cent,  yf^,  &c. 

328.  Percentage  is  the  result  obtained  by  finding  a 
certain  ger  cent  of  a  number. 

NOTE.  —  The  term  per  cent,  is  from  the  Latin  per,  by  and  centum, 
hundred. 

NOTATION    OF    PER    CENT. 

329.  The  Sign  of  Per  Cent  is  an  oblique  line 
between  two  ciphers  (%)  ;  .as  3$,  15$. 

NOTE.  —  The  sign  (%},  is  a  modification  of  the  sign  division  (-*-), 
the  denominator  100  being  understood.     Thus  5$  =-,-£u=  5-7-100. 

330.  Since  per  cent  denotes  a  certain  part  of  a  hundred, 
it  may  obviously  be  expressed  either  by  a  common  fraction, 
whose  denominator  is  100,  or  by  decimals,  as  seen  in  the 
following 

TABLE. 


3  per  cent  is  written  .03 

7  per  cent  "  .07 

10  per  cent  .10 

25  per  cent  "  .25 

50  per  cent  "  .50 

loo  per  cent  "  i.oo 

125  per  cent  "  1.25 

300  per  cent  "  3.00 


^  per  cent  is  written         .003 

|  per  cent  -0025 

f  per  cent  "              -0075 

f  per  cent  .006 

2^  per  cent  "               .025 

7£  per  cent  "               .074 

31  i  per  cent  -3125 

per  cent  "              1.125 


NOTES.  —  i.  Since  hundredths  occupy  two  decimal  places.it  follows 
hat  every  per  cent  requires,  at  least,  two  decimal  figures.  Hence, 

327.  What  do  the  termc  per  cent  and  rate  percent  denote  ?  328.  What  is  fi<—- 
ceutage?  Note.  Prom  what  are  the  terms  per  cent  and  percentage  derived? 
32q.  What  'is  the  sign  of  per  cent?  330.  How  may  per  cent  be  expressed? 
/Vote.  How  many  decimal  places  does  it  require?  Why?  If  the  tnven  per  cent 
Is  lc??  than  IP,  what  is  to  be  done?  What  is  100  i>cr  cent  of  a  nnmher? 


PERCENTAGE.  231 

If  the  given  per  cent  is  less  than  10,  a  cipher  must  be  prefixed  to 
the  figure  denoting  it.  Thus,  2  %  is  written  .02  ;  6  %,  .06,  etc. 

2.  A  hundred  per  cent  of  a  number  is  equal  to  the  number  itself; 
for  {ftll  is  equal  to  i.  Hence,  100  per  cent  is  commonly  written  i.oo. 

If  the  given  per  cent  is  100  or  over,  it  may  be  expressed  by  an 
integer,  a  mixed  number,  or  an  improper  fraction.  Thus,  125  pei 
cent  is  written  125  %,  1.25,  or  |gg.  Hence, 

331.  To  express  Per  Cent,  Decimally, 

Write  the  figures  denoting  the  per  cent  in  the  first  two 
places  on  the  right  of  the  decimal  point  ;  and  those  denoting 
parts  of  i  per  cent,  in  the  succeeding  places  toward  the  right. 

NOTES.  —  i.  When  a  given  part  of  i  per  cent  cannot  be  exactly 
expressed  by  one  or  two  decimal  figures,  it  is  generally  written  as  a 
common  fraction,  and  annexed  to  the  figures  expressing  the  integral 
per  cent.  Thus,  4},  c/0  is  written  .04^,  instead  of  .043333  +  • 

2.  In  expressing  per  cent,  when  the  decimal  point  is  used,  the 
words  per  cent  and  the  sign  (%)  must  be  omitted,  and  vice  versa. 
Thus.  .05  denotes  5  per  cent,  and  is  equal  to  TOO  or  -fa  ;  but  .05  per 
cent  or  .05$  denotes  TSn  of  T£u,  and  is  equal  to  TO&OTT  or 

Express  the  folloAving  per  cents,  decimally  : 

1.  2%,  6%,  8%,  14%,  20%,  35^,  60%,  12%. 

2.  80^,  101%,  104%,  150%,  210$,  300$. 
3-  i- 


332.   To  read  any  given  Per  Cent,  expressed  Decimally. 

Read  the  first  two  decimal  figures  as  per  cent;  and  those 
on  the  right  as  decimal  parts  of  i  per  cent. 

NOTE.  —  Parts  of  i  per  cent,  when  easily  reduced  to  a  common 
fraction,  are  often  read  as  such.  Thus  .105  is  read  10  and  a  half  per 
cent  ;  .0125  is  read  one  and  a  quarter  per  cent. 

Read  the  following  as  rates  per  cent  : 

4.  .05;  .07;  .09;  .045;  .0625;  .1875;  -I25;  -I65;  -27- 

5.  .10;  .17;  .0825;  .05125;  .33^;  .i6f;  .75375. 

6.  i.oo;  1.06;  2.50;  3.00;   1.125;  I-°725;  ^Sl- 


331.  How  express  per  cent,  decimally?  Note.  When  a  part  of  i  per  cent,  can- 
not be  exactly  expressed  by  one  or  two  decimal  figures,  how  is  it  commonly 
written  f  332.  How  read  a  given  per  cent,  expressed  decimally  ? 


232  PERCENTAGE. 

333.  To   change   a   given    Per  Cent  from   a    Decimal   to   a 

Common   Fraction. 

7.  Change  5^  to  a  common  fraction.     Ans.  TW— A- 

8.  Change  .045  to  a  common  fraction.  Ans.  -2§0. 

RULE. — Erase  the  decimal  point  or  sign  of  per  cent  (£), 
and  supply  the  required  denominator.  (Art.  179.) 

NOTE. — When  a  decimal  per  cent  is  reduced  to  a  common  fraction, 
then  to  its  lowest  terms,  this  fraction,  it  should  be  observed,  will 
express  an  equivalent  rate,  but  not  the  rate  per  cent. 

Change  the  following  per  cents  to  common  fractions : 

9.  5  percent;  io#;  4^;  20^;  25^;  50$;  75$. 
10.  6^  per  cent;  12^:  8£$;  33^;  62-^. 
n.  \  per  cent;  ££;  $#;  £#;  f#;  $#;  -&<£;  25^. 

334.  To  change  a  common  Fraction  to  an  equivalent  Per  Cent. 

12.  To  what  per  cent  is  £  of  a  number  equal  ? 

ANALYSIS. — Per  cent  denotes  hundredths.     The         I  —  i  .00  -f-  3 
question  then  is, how  is  i  reduced  to  hundredths?      r  oo_i. *—  ,^i 
Annexing  cipher's  to  the  numerator,  and  dividing 

by  the  denominator,  we  have  ^=1.00-5-3  or  .33 ^.  Hence,  the 

RULE. — Annex  two  ciphers  to  the  numerator,  and  divide 
ly  the  denominator.  (Art.  186.) 

13.  To  what  %  is  \  equal ?    i?    |?   -£?    f?    f?    |? 

14.  To  what  #  is  ^  equal  P-^?  ^?^?  •&?  T3T?  A? 

15.  To  what  ^  is  |  equal?    J?     |?    f?    1?    A?    «? 

335.  In  calculations  of    Percentage,  four  elements  01 
parts  are  to  be  considered,  viz. :  the  base,  the  rate  per  cent, 
b.Q  percentage,  and  the  amount. 

1.  The  base  is  the  number  on  which  the  percentage  is  calculated. 

2.  The  rntfe  /j«r  Cfnt    is  the  number  which   shows   liow  many 
hundredths  of  the  base  are  to  be  taken. 

333.  How  change  a  given  per  cent  from  a  decimal  to  a  common-  fraction  ? 
334.  How  change  a  common  fraction  to  an  equal  per  cent  ?  335.  How  xnany  parta 
»re  to  be  considered  in  ralcnMions  by  percentage? 


PE  11C  EXT  AGE.  233 

3.  The  percentage  is  the  number  obtained  by  taking  that  portion 
of  the  base  indicated  by  the  rate  per  cent. 

4.  The  amount  is  the  base  plus,  or  minus  the  percentage. 

The  relation  between  these  parts  is  such,  that  if  any  two  of  them 
are  given,  the  other  two  may  be  found. 

NOTES. — i.  The  term  amount,  it  will  be  observed,  is  here  employed 
in  a  modified  or  enlarged  sense,  as  in  algebra  and  other  departments 
of  mathematics.  Tliis  avoids  the  necessity  of  an  extra  rule  to  meet 
the  cases  in  which  the  final  result  is  less  than  the  base. 

2.  The  conditions  of  the  question  show  whether  the  percentage  is 
to  be  added  to,  or  subtracted  from  the  base  to  form  the  amount. 

3.  The  learner  should  be  careful  to  observe  the  distinction  between 
percentage  and  per  cent,  or  rate  per  cent. 

Percentage  is  properly  a  product,  of  which  the  given  per  cent  or 
rate  per  cent,  is  one  of  the  factors,  and  the  base  the  other.  This  care 
is  the  more  necessary  as  these  terms  are  often  used  indiscriminately. 

4.  The  terms  per  cent,  rate  per  cent,  and  rate,  are  commonly  used 
as  synonymous,  unless  otherwise  mentioned. 

PROBLEM    I. 

336.  To  find  the   Percentage,  the   Base   and    Rate  being 

given. 
Ex.  i.  What  is  5  per  cent  of  $600  ? 

ANALYSIS. — 5  per  cent  is  .05  ;  therefore  5  per  $600  B. 

cent  of  a  number  is  the  same  as  .05  times  that  .05  R. 

number.   Multiplying  the  base,  $600,  by  the  rate     ^ns.  $^o  oo  P 
.05,  and  pointing  off  the  product  as  in  multipli- 
cation of  decimals,  the  result  is  $30.    (Art.  191.)    Hence,  the 

RULE. — Multiply  the  base  by  the  rate,  expressed  deci- 
mally. 

FORMULA.    Percentage  —  Base  x  Rate. 

NOTES. — i.  When  the  rate  is  an  aliquot  part  of  100,  the  percentage 
may  be  found  by  taking  a  like  part  of  the  base.  Thus,  for  20  %, 
take  !;  ;  for  25%,  take  -\,  etc.  (Arts.  105,  270.) 

2.  When  the  bass  is  a  compound  number,  the  lower  denominations 
should  be  reduced  to  a  decimal  of  the  highest ;  or  the  higher  to  the 
lowest  denomination  mentioned ;  then  apply  the  rule. 

3.  Finding  a  per  cent  of  a  number  is  the  same  as  finding  a  frac- 
tional part  of  it,  etc.    The  pupil  is  recommended  to  review  with  care, 
Arts.  143,  165,  191 


Explain  them .    Wha  t  is  the  relation  of  these  parts »    The  difference  between  per- 
cen^ge  and  per  cent  ? 


234  PERCENTAGE. 

3.  3$  of  $807  ?  ii.  $$%  of  1000  men  ? 

4.  5$  of  216  bushels?  12.  io^$  of  1428  meters? 

5.  8%  of  282.5  yds.  ?  13.  50%  of  $1715-57  ? 

6.  4%  of  2 1 6  oxen  ?  1 4.  |$  of  £21.2? 

7-  5?%  of  150  yards  ?  15.  \%  of  500  liters? 

8.  16$  of  $72.40?  1 6.  f$  of  230  kilograms? 

9.  12$  of  840  Ibs.  ?  17.  100$  of  840  pounds  ? 
10.  14$  of  451  tons?  18.  200$  of  $500? 

19.  A  farmer  raised  875  bu.  of  corn,  and  sold  9$  of  it : 
how  many  bushels  did  he  sell  ? 

20.  The  gold  used  for  coinage  contains  10$  of  alloy: 
how  much  alloy  is  there  in  3^  pounds  of  standard  gold  ? 

21.  A  man  having  a  hogshead  of  cider,  lost  15$$  of  it 
by  leakage :  how  many  gallons  did  he  lose  ? 

22.  A  garrison  containing  4000  soldiers  lost  21$  of  them 
by  sickness  and  desertion  :  what  was  the  number  lost  ? 

23.  A  grocer  having  1925  pounds  of  sugar,  sold  \2\  per 
cent  of  it :  how  many  pounds  did  he  sell  ? 

ANALYSIS. — 12.3  %  is  \-  of  1005,  and  i<x>%  of  OPERATIOH. 

a  number  is  equal  to  the  number  itself ;  there-  8)1925  Ibs. 

fore  TI\  per  cent  of  a  number  is  equal  to  \  of     Ans.     240.62  < 
that  number,  and  ^  of  1925  Ibf ...  is  240%  Ibs.     In 
the  operation  we  take  ^  of  the  base. 

Solve  the  next  9  examples  by  aliquot  parts : 

24.  Find  25$  of  $860.  26.  1 2^$  of  258  meters. 

25.  10$  of  1572  pounds.        27.  20$  of  580  liters. 

28.  A  drover' taking  2320  sheep  to  market,  lost  25$  of 
them  by  a  railroad  accident :  how  many  did  he  lose  ? 

29.  A  farmer  raised  468  bu.  of  corn,  and  33^$  as  many 
oats  as  corn :  how  many  bushels  of  oats  did  he  raise  ? 

30.  A  young  man  having  a  salary  of  $1850  a  year,  spent 
50  per  cent  of  it :  what  were  his  annual  expenses  ? 

31.  What  is  334$  of  1728  cu.  feet  of  wood  ? 

32.  What  is  12^  per  cent  of  £16,  8s.  ? 

336.  How  find  the  percpnta<ro  vrhon  the  base  and'  rate  arc  piveti  ?    When  thf 
rate  is  an  aliquot  part  of  100,  how  proceed  ?    When  a  compound  number  f 


PERCENTAGE.  235 

PROBLEM    II. 
337.    To  find  the  Amount,  the  Base  and  Rate  being  given, 

1.  A  commenced  business  Atith  $1500  capital,  and  laiJ 
up  S%  the  first  year:  what  amount  was  he  then  worth  ? 

ANALYSIS. — Since  he  laid  up   8fc,  he  was  OPERATION. 

worth  his  capital,  $1500,  plus  8f0    of  itself.  $1500    B. 

But  his  capital  is  100%  or  i  time  iieolf ;  and  1.08,  i  H  Ii. 

100% +8%=io8%   or  1.08;  therefore  he  was  120.00 

worth  i. 08  times  $1500.     Now  $1500  x  1.08  =;  1500 

$1620.    We  multiply  the  base  by  i  plus  the  $1620  oo  Am't 
given  rate,  expressed  decimally.    (Art.  191.) 

2.  B  commenced   business  with   $1800    capital,  and 
squandered  6%  the  first  year:  what  was  he  then  worth"? 

ANALYSIS. — As  B  squandered  6%,  he  was  $1800  B. 

worth  his  capital  $1800,    minus  6%  of  itself.  .04  R. 

But  his  capital  is  100%  or  i  time  itself;  and  •7700 

100%  —6%  =94%  or  .94.    Therefore  he  had  .94  T62oo 

times  $1800  ;  and  $1800  x  .94=$i692.     Here  we  v—                .      , 

multiply  the  base  by  i  minus  the  given  rate,  $l692-°°  Am  t 
expressed  decimally.     (Art.  191.)    Hence,  the 

EULE. — Multiply  the  base  by  i  plus  or  minus  the  rate,  as 
the  case  may  require.  The  result  will  be  the  amount. 

FORMULA.    Amount  =.  Base  x  (i  ±  Rate). 

NOTE. — i.  The  character  (  ±  )  is  called  the  double  or  amJnguous 
sign.  Thus,  the  expression  $5  ±  $3  signifies  that  $3  is  to  be  added 
to  or  subtracted  from  $5,  as  the  case  may  require,  and  is  read, 
"  $5  plus  or  minus  $3." 

2.  The  rule  is  based  upon  the  axiom  that  the  whole  is  equal  to 
the  sum  of  all  its  parts. 

3.  When,  by  the  conditions  of  the  question,  the  amount  is  to  be 
greater  than  the  base,  the  multiplier  is  i  plus  the  rate  ;  when  the 
amount  is  to  be  less  than  the  base,  the  multiplier  is  I   mi;ius  the 
rate. 

337.  How  find  the  amount  when  the  base  and  rate  are  given  ?  338.  How  el*e 
\»  the  amount  found,  when  the  base  aud  ratq  arc  given  ? 


236  PERCENTAGE. 

338.  When  the  base  and  rate  are  given,  the  amount  rujv 
also  be  obtained  by  first  finding  the  percentage,  then  adding 
it  to  or  subtracting  it  from  the  base.  (Art.  336.) 

3.  0  and  D  have  1000  sheep  apiece;  if  C  adds  15%  to 
his  flock,  and  D  sells   1 2%  of  his,  how  many  sheep  will 
each  have  ? 

4.  A  merchant  having  $2150.38  in  bank,  deposited 
1%  more :  what  amount  had  he  then  in  bank  ? 

5.  If  you  have  $3000  in  railroad  stock,  and  sell  $%  of  it, 
what  amount  of  stock  will  you  then  have  ? 

6.  The  cotton  crop  of  a  planter  last  year  was  450  bales; 
this  year  it  is  12  per  cent  more :  what  is  his  present  crop  ? 

7.  An  oil  well  producing  2375  gallons  a  day,  loses  15$ 
of  it  by  leakage :  what  amount  per  day  is  saved  ? 

8.  A  gardener  having  1640  melons  in  his  field,  lost  20^' 
of  them  in  a  single  night:  what  number  did  he  have 
left? 

9.  A  man  paid  $420  for  his  horses,  and  12%  more  for 
his  carriage :  what  was  the  amount  paid  for  the  carriage  ? 

10.  A  man  being  asked  how  many  geese  and  turkeys  he 
had,  replied  that  he  had  150  geese;  and  the  number  of 
turkeys  was  14$  less:  how  many  turkeys  had  he? 

it.  A  fruit  grower  having  sent  2500  baskets  of  peaches 
to  New  York,  found  9%  of  them  had  decayed,  and  sold  the 
balance  for  62  cts.  a  basket:  what  did  he  receive  for  his 
peaches  ? 

12.  A  Floridian  having  4560  oranges,  bought  25%  more, 
and  sold  the  whole  at  4  cts.  each :  what  did  he  receive 
for  them  ? 

13.  If  a  man's  income  is  $7235  a  year,  and  he  spends 
33  J$  of  it,  what  amount  will  he  lay  up  ? 

14.  A  man  bought  a  house  for  $8500,  and  sold  it  for 
20$  more  than  he  gave  :  what  did  he  receive  for  it  ? 

15.  A  merchant  bought  a  bill  of  goods  for  $10000,  and 
s  >ld  them  at  a  loss  of  2\%\  what  did  he  receive  ? 


PEBCENTAGE.  237 

PROBLEM     III. 

339.  To  find  the  Rate,  the  Base  and  Percentage  being  given ; 
Or,  to  find  what  Per  Cent  one  number  is  of  another. 

i.  A  clerk's  salary,  being  $1500  a  year,  was  raised  $250*, 
what  rate  was  the  increase  ? 

ANALYSIS. — Iii   this  example   $150x5  is   the  OPERATION. 

base,  and  $250  the  percentage.     The  question      I5oo)$25o.oo  P. 
then  is  this:  $250  is  what  per  cant  of  $1500?  Ans 

Now  $250  is  i^5,,0-,,  of  $1500;  and  $25o-j-$i5oo 
=.16666,  etc.,  or  16$$.      The  first  two  decimal  figures  denote  the 
per  cent ;  the  others,  parts  of  i#.      (Arts.  331,  2.)     Hence,  the 

EULE. — Divide  the  percentage  ly  the  base. 

FORMULA.  Rate  =  Percentage  -j-  Base. 

NOTES.— i.  This  prob.  is  the  same  as  finding  what  part  one  number 
is  of  another,  then  changing  the  common  fraction  to  hundredths. 
(Arts.  173,  186,  334.) 

It  is  based  upon  the  principle  that  percentage  is  a  product  ot 
which  the  base  is  a  factor,  and  that  dividing  a  product  by  one  of  its 
factors  will  give  the  other  factor.  (Art.  93.) 

2.  The  number  denoting  the  base  is  always  preceded  by  the  word 
of,  which  distinguishes  it  from  the  percentage. 

3.  The  given  numbers  must  be  reduced  to  the  same  denomination-, 
and  if  there  is  a  remainder  after  two  decimal  figures  arc  obtained, 
place  it  over  the  divisor  and  annex  it  to  the  quotient. 

2.  What  %  of  15  is  2  ?          6.  What  %  of  £8  are  1 53.  ? 

3.  What  %•<$  $20  are  $5  ?    7.  What  %  of  56  gals,  are  7  qts.? 

4.  What  %  of  48  is  1 6  ?        8.  What  %  are  5  dimes  of  $5  ? 

5.  What  %  of  $5  are  75  cts.?  9.  What  %  of  f  ton  is  \  ton  ? 

10.  The  standard  for  gold  and  silver  coin  in  the  U.  S. 
is  9  parts  pure  metal  and  i  part  alloy :  what  %  is  the  alloy :' 

11.  From  a  hogshead  of  molasses  15  gals,  leaked  out 
what  per  cent  was  the  leakage  ? 

12.  A  grocer  having  560  bbls.  of  flour,  sold  \  of  it. 
what  per  cent  of  his  flour  did  he  sell  ? 

13.  A  horse  and  buggy  are  worth  $475;  the  buggy  is 
worth  $i  10 ;  what  %  is  that  of  the  value  of  the  horse  ? 

330.  How  find  the  rate,  when  the  base  and  percentage  are  given  ?  To  what  la 
this  problem  equivalent  ?  JVbre.  Upon  what  is  it  based  ? 


238  PEBCENTAGE. 

PROBLEM    IV. 
340.  To  find  the  Base,  the  Percentage  and  Rate  being  given. 

1.  A  father  gave  his  son  $30  as  a  birthday  present, 
which  was  6%  of  the  sum  he  gave  his  daughter:  how 
much  did  he  give  his  daughter  ? 

ANALYSIS. — The  percentage  $30  is  the  product  of        OPERATION. 
the  base  into  .06  the  rate;  therefore  $30-^.06  is  the      .o6)$3o.oo 
other  factor  or  bass  ;  and  $30-7- .06 =$500,  the  sum  he      ^jsTlfecoo 
gave  his  daughter.  (Art.  193,  n.) 

Or,  since  $30  is  6  %  of  a  number,  i  %  of  that  number  must  be  \  of 
$30,  which  is  $5  ;  and  100%  is  100  times  $5  or  $500. 

It  is  more  concise,  and  therefore  preferable,  to  divide  the  per- 
centage by  the  rate  expressed  decimally  ;  then  point  off  the  quotient 
as  in  division  of  decimals.  (Art.  193.)  Hence,  the 

RULE. — Divide  the  percentage  ty  the  rate,  expressed 
decimally. 

FORMULA.    Base  =  Percentage  -f-  Rate. 

NOTES.— i.  This  problem  is  the  same  as  finding  a  number  when 
a  given  per  cent  or  a  fractional  part  of  it  is  given.  (Arts.  174,  334.) 

2.  The  rule,  like  the  preceding,  is  based  upon  the  principle  that 
percentage  is  a  product,  and  the  rate  one  of  its  factors.    (Art.  335,  n.) 

3.  Since  the  percentage  is  the  same  part  of  the  base  as  the  rate  is 
of  100,  when  the  rate  is  an  aliquot  part  of  100,  the  operation  will  to 
shortened  by  using  this  aliquot  part  as  the  divisor. 

2.  40  is  i2^Nof  what  number? 
SOLUTION. — \2\%=\  and  40-74=40x8=320.  An*. 

3.  20=5$  of  what  number?  Ans.  400. 

4.  15  bushels=:6%  of  what  number? 


5.  $29  =  8%  of  what? 

6.  45  tons=25%  of  what? 

7.  £150=133^  of  what? 

8.  37-5=6^  of  what? 

9.  45  francs  =  1 2\%  of  what? 


10.  40—  \%  of  what  ? 

11.  50  c,\&.  —  \%  of  what  ? 

12.  $ioo=|^  of  what? 

13.  $35.20—3$  of  what? 

14.  68  yds.=  125$  of  what  ? 


140.  How  flud  the  base,  when  the  percentage  and  rate  are  given  f    Note.  Upon 
what  does  the  rule  depend  f    When  the  rate  is  an  aliquot  part  of  100,  how  proceed  f 


PERCENTAGE.  2 

15.  z%  of  $150  is  (>%  of  what  sum? 

1  6.  12%  of  500  is  60%  of  what  number? 

17.  A  paid  a  school  tax  of  $50,  which  was  i%  on  the 
valuation  of  his  property  :  what  was  the  valuation  ? 

1  8.  B  saves  31^$  of  his  income,  uud  lays  up  $600:  what 
is  his  income  ? 

19.  A  general  lost  i6£%  of  his  army,  315  killed,  no 
prisoners,  and  70  deserted:  how  many  men  had  he? 

20.  According  to  the  bills  of  mortality,  a  city  loses  450 
persons  a  month,  and  the  number  of  deaths  a  year  is  \\% 
of  ics  population  :  what  is  its  population  ? 

PROBLEM    V. 

341.  To  find  the  Base,  the  Amount  and  Rate  being  given. 
i.  A  manufacturer  sold  a  carriage  for  $633,  which  was 
more  than  it  cost  him  :  what  was  the  cost  ? 


ANALYSIS.—  The  amount  received  $633,  is  OPERATION. 

equal  to  the  cost  or  base  plus  $\%   of  itself.      1.055)^633.000 
Now  the  cost  is  100%   or  i  time  itself,  and      _Ans.  $600 

100%  +5^  =  1.  05^;  hence  $633  equals  1.05^ 

times  the  cost  of  the  carriage.  The  question  now  is:  633  is  105-^$ 
or  1.05  i  times  what  number?  If  633  is  1.05^  times  a  certain  num- 
ber, once  that  number  is  equal  to  as  many  units  as  1.05!  is  contained 
times  in  633  ;  and  633-7-1.055=600.  Therefore  the  cost  was  $600. 

2.  A  lady  sold  her  piano  for  $628.  25,  which  was  12-^ 
less  than  it  cost  her:  what  was  the  cost? 

ANALYSIS.  —  There  being  a  loss  in  this  case,          OPERATION. 
the  amount  received,  $628.25,  equals  the  cost 


or  base  minus  \z\%  of  itself.    But  the  cost  is     Ans.        $7  18 
100^7  or  i  time  itself,  and  i<X)%  —  i.2\  %  =  .%•}}  ; 
hence  $628.25  equals  .875  times  the  cost.     Now  if  $628.25  equals 
.87^  times  the  cost,  once  the  cost  must  be  as  many  dollars  as  .87-}-  is 
contained  times  in  $628.25,  or  $718.     Hence,  the 

RULE.  —  Divide  the  amount  by  i  plus  or  minus  the  rate,. 
as  the  case  may  require. 

FORMULA.     Base  =  Amount  -f-  (i  ±  Rate}. 

341.  How  find  the  base,  the  amount  and  rate  being  given?    Note.  Upon  what 


240  PEBCENTAGE. 

NOTES. — i.  This  problem  is  the  same  as  finding  a  number  which 
is  a  given  per  cent  greater  or  less  than  a  given  number. 

2.  The  rule  depends  upon  the  principle  that  the  amount  is  a  pro- 
duct of  which  the  base  is  one  of  the  factors,  and  i  plus  or  minu*  the 
rate,  the  other. 

3.  Thp  nature  of  the  question  shows  whether  i  is  to  be  increased 
or  diminisfied  by  the  rate,  to  form  the  divisor. 

4.  When  the  rate  is  an  aliquot  part  of  100,  the  operation  is  often 
shortened  by  expressing  it  as  a  common  fraction.     Thus25$>=:i; 
and  i  or  }  +  ^=f,  etc. 

3.  What  number  is  8%  of  itself  less  than  351  ?  A.   325. 

4.  "What  number  is  $\%  of  itself  more  than  378  ?  A.  400. 

5.  What  number  diminished  33^  of  itself  will  equal 
539i  ? 

6.  2275  is  25%  more  than  what  number? 

7.  £  is  i2\%  more  than  what  number? 
ANALYSIS. — i2i$=i ;  and  \ -5-i£=$ -4-t=f §•  or  f-  An** 

8.  -f  is  10%  less  than  what  number  ? 

9.  A  owns  \  of  a  ship,  which  is  1 6f  %  less  than  B's  part : 
what  part  does  B  own  ? 

10.  A  garrison  which  had  lost  28^  of  its  men,  had  3726 
left :  how  many  had  it  at  first  ? 

1 1.  A  merchant  drew  a  check  for  $45 60,  which  was  25$ 
more  than  he  had  in  bank  :  how  much  had  he  011  deposit  ? 

12.  The  population  of  a  certain  place  is  8250,  which  is 
20%  more  than  it  was  5  years  ago :  how  much  was  it  then  ? 

13.  A  man  lays  up  $2010,  which  is  40%  less  than  his  in- 
come :  what  is  his  income  ? 

14.  A  drover  lost  10%  of  his  sheep  by  disease,  15$  were 
stolen,  and  he  had  171  left :  how  many  had  he  at  first  ? 

15.  The  attendance  of  a  certain  school  is  370,  and  7 1 
of  the  pupils  are  absent :  what  is  the  number  on  register? 

1 6.  An  army  having  lost  iofc  in  battle,  now  contains 
5220  men  :  what  was  its  original  force  ? 

docs  this  rule  depend  ?  How  determine  whether  i  is  to  be  increased  or  dimin- 
ished by  the  rate  ?  When  the  rate  is  au  aliquot  part,  how  proceed?  To  what  U 
this  problem  equivalent  1 


APPLICATIONS   OF   PEKCENTAGE. 

342.  The  Principles  of  Percentage  are  applied 
to  two  important  classes  of  problems: 

First.  Those  in  which  time  is  one  of  the  elements  o; 
calculation ;  as,  Interest,  Discount,  etc. 

Second.  Those  which  are  independent  of  time ;  as,  Com- 
mission, Brokerage,  and  Profit  or  Loss. 

COMMISSION    AND    BROKERAGE. 

343.  Commission  is  an  allowance-  made  to  agents, 
collectors,  brokers,  etc.,  for  the  transaction  of  business. 

JZroJierage  is  Commission  paid  a  broker. 

NOTES. — i.  An  Agent  is  one  who  transacts  business  for  another, 
and  is  often  called  a  Commission  Merchant,  Factor,  or  Correspondent. 

2.  A  Collector  is  one  who  collects  debts,  taxes,  duties,  etc. 

3.  A  Broker  is  one  who  buys  and  sells  gold,  stocks,  bills  of  ex- 
change,  etc.     Brokers  are  commonly  designated  by  the  department 
of  business  in  which  they  are  engaged ;  as,  Stock-brokers,  Exchange- 
brokers,  Note-brokers,  Merchandise-brokers,  Real-estate-brokers,  etc. 

4.  Goods  sent  to  an  agent  to  sell,  are  called  a  consignment ;  th« 
person  to  whom  they  are  sent,  the  consignee  ;  and  the  person  send- 
ing them  the  consignor. 

344.  Commission  and  Brokerage  are  computed   at  a 
certain  per  cent  of  the  amount   of  business   transacted. 
Hence,  the  operations  are  precisely  the  same  as  those  in 
Percentage.     That  is, 

The  sales  of  an  agent,  the  sum  collected  or  invested  by 
lim,  are  the  base. 

The  per  cent  for  services,  the  rate. 
The  commission,  the  percentage. 
The  sales,  etc., plus  or  minus  the  commission,  theltmount. 

342.  To  what  two  elapses  of  problems  are  the  principles  of  percentage  applied  T 
•543.  What  is  commission?  Brokerage?  JVofe.  An  agent?  What  called?  A 
collector?  Broker?  344.  How  are  commission  and  brokerage  computed ?  What 
IB  the  base  ?  The  rate  ?  The  percentage  ?  The  amount  ?  Note.  The  net  proceeds  f 


242  COMMISSION     AND     BROKERAGE. 

NOTES. — i.  The  rate  of  commission  and  brokerage  varies.  Com- 
mission merchants  usually  charge  about  2\  per  cent  for  selling 
goods,  and  2^  per  cent  additional  for  guaranteeing  the  payment 
Stock-brokers  usually  charge  £  per  cent  on  the  par  value  of  stocks, 
without  regard  to  their  market  value. 

2.  The  net  proceeds  of  a  business  transaction,  are  the  gross  amount 
of  sales,  etc.,  minus  the  commission  and  other  charges. 

345.    To  flnd  the   Commission,  the  Sales   and   the   Rate 
being  given. 

Multiply  the  sales  by  the  rate.     (Problem  I,  Percentage.) 

NOTES. — i.  When  the  amount  of  sales,  etc.,  and  the  commission 
are  known,  the  net  proceeds  are  found  by  subtracting  the  commission 
from  the  amount  of  sales.  Conversely, 

2.  When  the  net  proceeds  and  commission  are  known,  the  amount 
of  sales,  etc.,  is  found  by  adding  the  commission  to  the  net  proceeds. 

3.  When  both  the  amount  of  sales,  etc.,  and  the  net  proceeds  are 
known,  the  commission  is  found  l\y  subtracting  the  net  proceeds 
from  the  amount  of  sales. 

4.  In  the  examples  relating  to  stocks,  a  share  is  considered  $100, 
unless  otherwise  mentioned.     (Ex.  2.) 

(For  methods  of  analysis  and  of  deducing  the  rules  in  Commission, 
Profit  or  Loss,  etc.,  the  learner  is  referred  to  the  corresponding 
Problems  in  Percentage.) 

1.  A  merchant  sold  a  consignment  of  cloths  for  $358: 
what  was  his  commission  at  2\  per  cent  ? 

SOLUTION. — Commission =$358  (sales)  x  .025  (rate)— $8.95.  Ans. 

2.  A  broker  sold  39  shares  of  bank  stock:  what  was  his 
brokerage,  at  \  per  cent  ? 

SOLUTION. — 39  shares =$3900;  and  $3900  x  .005  =  119.500.  Am. 

3.  Sold  a  consignment  of  tobacco  for  $958.25  :  what  was 
my  commission  at  3^  ? 

4.  A  man  collected  bills  amounting  to  $11268.45,  and 
charged  zY/'c'-  w^at  was  his  commission;  and  how  much 

did  he  pay  his  employer  ? 

• 

345.  How  find  the  ooimnisfion  or  brokerage,  when  the  sales  and  the  rate  arc 
2ft'*,  How  find  the  r.et  proceeds  ? 


COMMISSION     AND     BROKERAGE. 

—  5.  A  commission  merchant  sold  a  consignment  of  goods 
for  $4561,  and  charged  2\%  commission,  and  •$%  for 
guaranteeing  the  payment :  what  were  the  net  proceeds  ? 
—••6.  An  agent  sold  1530  Ibs.  of  maple  sugar  at  i6f  cts., 
for  which  he  received  z\%  commission:  what  were  the 
net  proceeds,  allowing  $7.50  for  freight,  and  $3.10  for 
storage  ? 

346.  To  find  the  Rate,  the  Saies  and  the  Commission 

being  given. 
Divide  the  commission  by  the  sales.  (Prob.  III.  Percentage.) 

7.  An  auctioneer  sold  goods  amounting  to  $2240,  for 
which   he  charged  $53.20  commission :    what  per  cent 
was  that  ? 

SOLUTION. — Per  cent  =  $53.20  (com.)  -f-  $2240  (sales)  =  .02375,  or 
2j  per  cent.  (Art.  339,  331,  n.) 

8.  A  broker  charged  $19   for  selling  $3800  railroad 
stock :  what  per  cent  was  the  brokerage  ? 

9.  Received  $350  for  selling  a  consignment  of  hops 
amounting  to  $7000:    what  per  cent  was  my  commis- 
sion ? 

10.  An  administrator  received  $118.05  f°r  settling  an 
estate  of  $19675  :  what  per  cent  was  his  commission  ? 

347.  To  find   the  Sales,   the   Commission   and   the   Rate 

being  given. 

Divide  the  commission  by  the  rate.     (Prob.  IV,  Per  ct.) 

n.  An  agent  charged  2%  for  selling  a  quantity  of 
muslins,  and  received  $93.50  commission :  what  was  the 
amount  of  his  sales  ? 

SOLUTION.— Sales= $93. 50  (com.)-?- .02  (rate)=$4675.  Am. 

346.  How  find  the  per  cent  commission,  when  the  sales  and  the  commission 
are  given  ?  347.  How  find  the  amount  of  sales,  when  the  commission  and  the 
rate  arc  ffivon  ? 


244  COMMISSION     AND     BKOKEBAGE 

12.  Received  $45  brokerage  for  selling  stocks,  which 
was  \%  of  what  was  sold :  what  was  the  amount  of  stocks 
sold? 

13.  A  commission  merchant  charging  2\%  commission, 
and  2  \%  for  guaranteeing  the  payment,  received  $210.60  for 
selling  a  cargo  of  grain :  what  were  the  amount  of  sales, 
and  the  net  proceeds  ?  — 

14.  A  district  collector  received  $67.50  for  collecting  a 
school  tax,  which  was  $\%  commission :  how  much  did  he 
collect,  and  how  much  pay  the  treasurer  ? 

15.  An  auctioneer  received  $135  for  selling  a  house, 
which  was  i$%:   for  what  did  the  house  sell;  and  how 
much  did  the  owner  receive  ? 

348.    To  find  the  Sales,  the  net  proceeds  and  per  cent 
commission  being  given. 

Divide  the  net  proceeds  by  i  minus  the  rate.     (Prob.  V.) 

1 6.  An  agent  sold  a  consignment  of  goods  at  z\%  com- 
mission, and  the  net  proceeds  remitted  the  owner  were 
$3381.30  :  what  was  the  amount  of.  sales? 

SOLUTION.— Sales=$338i. 30  (net  p.)-7-.g75  (i— rate) =$3468.  Ana. 

17.  A  tax  receiver  charged  5%  commission,  and  paid 
$4845  net  proceeds  into  the  town  treasury :  what  was  the 
amount  collected? 

NOTE. — In  this  and  similar  examples,  the  pupil  should  observe 
that  the  base  or  sum  on  which  commission  is  to  be  computed  is  the 
sum  collected,  and  not  the  sum  paid  over.  If  it  were  the  latter,  the 
agent  would  have  to  collect  his  own  commission,  at  his  own  expense, 
and  his  rate  of  commission  would  not  be  rSir.  but  T£S.  In  the  col- 
lection of  $100,000,  this  would  cause  an  error  of  more  than  $350. 

1 8.  After  retaining  z\%  for  selling  a  consignment  of 
flour,  my  agent  paid  me  $6664 :  required  the  amount  of 
sales,  and  his  commission. 

348.  How  find  the  sales,  etc.,  the  net  proceeds  and  the  per  cent  commission 
being  given  f 


COMMISSION     AND     BROKERAGE.  245 

— rg.  After  deducting  \\%  for  brokerage,  and  $45.28  for 
advertising  a  house,  a  broker  sent  the  owner  $15250:  for 
what  did  the  house  sell  ? 

20.  An  administrator  of  an  estate  paid  the  heirs  $25686, 
""charging  z\%  commission,  and  $350  for  other  expenses: 

what  was  the  gross  amount  collected  ? 

349.  To  find  the  Sum  Invested,  the  sum  remitted  and  the 
per  cent  commission  being  given. 

21.  A  manufacturer  sent  his  agent  $3502  to  invest  i:i 
wool,  after  deducting  his  commission  of  3% :  what  sum  di<  i 
he  invest  ? 

ANALYSIS. — The  sum  remitted  $3502,  includes  both  the  sum  in- 
vested and  the  commission.  But  the  sum  invested  is  ioofc  of  itself, 
andioo#+3#  (the  commission)  -  103$.  The  question  now  is : 
$3502  is  103%  of  what  number?  $35O2-^i.O3=$34oo,  the  sum  in- 
vested. (Art.  340,  «.)  Hence,  the 

RULE. — Divide  the  sum  remitted  by  i  plus  the  per  cent 
commission.  (Prob.  V,  Per  ct.) 

NOTE. — The  learner  will  observe  that  the  base  in  this  and  sim 
ilar  examples  is  the  sum  invested,  and  not  the  sum  remitted.  If  it 
were  the  latter,  the  agent  would  receive  commission  on  his  commis- 
sion, which  is  manifestly  unjust. 

22.  A  teacher  remitted  to  an  agent  $3131.18  to  be  laid 
out  in  philosophical  apparatus,  after  deducting  4^'  com- 
mission :    how  much   did    the    agent    lay  out  in  appa- 
ratus ? 

23.  If  I  remit  my  agent  $2516  to  purchase  books,  after 
deducting  4$  commission,  how  much  does  he  lay  out  in 
books  ? 

24.  Remitted  $50000  to  a  broker  to  be  invested  in  city 
property,  after  deducting  \\c/r  for  his  services:  how  much 
did  he  invest,  and  what  was  his  commission  ? 


340.  How  find  the  sum  invested,  the  sum  remitted  and  the  per  cent  commis- 
sion being  given  ? 


246 


ACCOUNT     OF     SALES. 


ACCOUNT    OF    SALES. 

350.  An  Account  of  Sales  is  a  written  statement, 
made  by  a  commission  merchant  to  a  consignor,  contain- 
ing the  prices  of  the  goods  sold,  the  expenses,  and  the  net 
proceeds.  The  usual  form  is  the  following : 

Sales  of  Grain  on  acc't  of  E.  D.  BARKER,  ESQ.,  Chicago. 


DATE. 

BUTEB. 

DESCRIPTION. 

BUSHELS. 

PRICK. 

EXTENSION. 

IS?!. 

April   3 

J.  Hoyt, 
A.  Woodruff, 
Hecker  &  Co., 

Winter  wheat, 
Spring       " 
Corn, 

565 
870 
1610 

@,  $2.10 
@      1-95 
@      1.05 

$1186.50 
1696.50 
1690.50 

Gross  amount, 


Charges. 


Freight  on  3045  bu.,  at  20  cts., 
Cartage  "  ''       $15.30, 

Storage  "         38.75, 

Commission,  2\  % , 


$609.00 

1530 

38.75 

"4-34 


Net  proceeds, 


$4573.50 


$777-39 
$3796.11 


YORK,  July  sth,  1871. 


J.  HENDERSON  &  Co. 


Ex.  25.  Make  out  an  Account  of  Sales  of  the  following: 

James  Penfield,  of  Philadelphia,  sold  on  account  of 
J.  Hamilton,  of  Cincinnati,  300  bbls.  of  pork  to  W.  Gerard 
&  Co.,  at  $27;  1150  hams,  at  $1.75,  to  J.  Eamsey;  87*5 
kegs  of  lard,  each  containing  50  lb.,  at  8  cts.,  to  Henry 
Parker,  and  750  lb.  of  cheese,  at  10  cts.,  to  Thomas  Young. 

Paid  freight,  $65.30;  cartage,  $15.25  ;  insurance,  $6.45  ; 
commission,  at  2%.  What  were  the  net  proceeds  ? 

26.  Samuel  Barret,  of  New  Orleans,  sold  on  account  of 
James  Field,  of  St.  Louis,  85  bales  cotton,  at  $96.50; 
63  barrels  of  sugar,  at  $48.25  ;  37  bis.  molasses,  at  $35. 

Paid  freight,  $45.50;  insurance,  $15;  storage,  $35.50; 
>-\%-  What  were  the  net  proceeds  ? 

350.  What  is  an  account  of  sales  f 


PROFIT     AND     LOSS.  247 


PROFIT    AND    LOSS. 

351.  Profit  and  Loss  are  the  sums  gained  or  lost  in 
business  transactions.  They  are  computed  at  a  certain 
per  cent  of  the  cost  or  sum  invested,  and  the  operations 
are  the  same  as  those  in  Percentage  and  Commission. 

The  Cost  or  sum  invested  is  the  Base  ; 

The  Per  cent  profit  or  loss,  the  Rate  ; 

The  Profit  or  Loss,  the  Percentage  ; 

The  Selling  Price,  that  is,  the  cost  plus  or  minus  tho 
profit  or  loss,  the  Amount. 

352.  To  find  the  Profit  or  Loss,  the  Cost  and  the  Per  Cent 
Profit  or  Loss  being  given. 

Multiply  the  cost  by  the  rate.     (Problem  1,  Per  ct.) 

NOTE. — When  the  per  cent  is  an  aliquot  part  of  100,  it  is  gener- 
ally shorter,  and  therefore  preferable  to  use  the  fraction.  (Art.  336,  n.) 

1.  A  man  paid  $250  for  a  horse,  and  sold  it  at  15$ 
profit:  how  much  did  he  gain?  Ans.  $37.50. 

2.  A  man  paid  $450  for  a  building  lot,  and  sold  at  a 
loss  of  n$:  how  much  did  he  lose?  Ans.  $49.50. 

3.  Paid  $185  for  a  buggy,  and  sold  it  12%  less  than  cost: 
what  was  the  loss  ? 

4.  Paid  $110  for  a  pair  of  oxen,  and  sold  them  at  20% 
advance :  what  was  the  profit  ? 

5.  A  lad  gave  87^  cts.  for  a  knife,  and  sold  it  at  10% 
below  cost :  how  much  did  he  lose  ? 

6.  Bought  a  watch  for  $83^,  and  sold  it  at  a  loss  of  20^: 
what  was  the  loss  ? 

7.  Bought  a  pair  of  skates  for  $4.20,  and  sold  them  at 
advance :  required  the  gain  ? 


351.  What  are  profit  and  loss?  How  reckoned?  To  what  does  the  cost  or 
sum  invested  answer  ?  The  per  cent  profit  or  loss  ?  The  profit  or  loss*  ?  The 
selling  price?  352.  How  find  the  profit  or  loss,  when  the  cost  and  per  cent  are 
Note.  When  the  per  cent  ie  an  aliquot  part  of  100,  how  proceed? 


248  PKOFIT     AND     LOSS. 


353.    To  find  the  Selling  Price,  the  Cost  and   Per  Cent 
Profit  or  Loss  being  given. 

Multiply  the  cost  by  i  plus  or  minus  the  per  cent.  (Prob. 
II.  Percentage.) 

NOTE. — When  the  cost  and  per  cent  profit  or  loss  are  given,  the 
telling  price  may  also  be  found  by  first  finding  the  profit  or  loss ; 
then  add  it  to  or  subtract  it  from  the  cost.  (Art.  338.) 

8.  A  man  paid  $300  for  a  house  lot :  for  what  must  he 
sell  it  to  gain  20$!  ? 

ANALYSIS. — To  gain  20%,  he  must  sell  it  for  the  cost  plus  20  (>r. 
That  is,  selling  pr.=f  300  (cost)  x  1.20  (i  +  20$)= $360.  Ana. 

9.  A  farmer  paid  $250  for  a  pair  of  oxen:   for  how 
much  must  he  sell  them  to  lose  15$? 

Selling  pr.=$25o  (cost)  x  .85  (1  —  15$) =$2 12.50.  Am. 

10.  A  and  B  commenced  business  with  $2500  apiece. 
A  adds  11%  to  his  capital  during  the  first  six  months,  and 
B  loses  \i%  of  his:  what  amount  is  each  then  worth  ? 

n.  A  merchant  paid  $378  for  a  lot  of  silks,  and  sold 
them  at  20%  profit :  what  did  he  get  for  the  goods  ? 

12.  If  a  man  pays  $2750  for  a  house,  for  how  much 
must  he  sell  it  to  gain  1%  ? 

13.  If  a  man  starts  in  business  with  a  capital  of  $8000, 
and  makes  19$  clear,  how  much  will  he  have  at  the  close 
of  the  year  ? 

14.  If  a  merchant  pays  15  cts.  a  yard  for  muslin,  how 
must  he  sell  it  to  lose  25%? 

15.  Bought  gloves  at  $15  a  dozen:  how  must  I  sell 
them  a  pair,  to  lose  20%? 

1 6.  Paid  $25  per  dozen  for  pocket  handkerchiefs:  for 
what  must  I  sell  them  apiece  to  make  33^  per  cent? 


353.  How  find  the  selling  price,  when  the  cost  and  per  cent  profit  or  loss  aw 
(flven.    Note.  How  else  may  the  selling  price  be  found  ? 


PEOFIT     AND     LOSS.  249 

17.  Paid  $196  for  a  piece  of  silk  containing  50  yds.: 
how  must  I  sell  it  per  yard  to  gain  25$  ? 

18.  Bought  a  house  for  $3850:  how  must  I  sell  it  to 
make  12%%? 

19.  A  speculator  invested  $14000  in  flour,  and  sold  at  a 
loss  of  8-fyc :  what  did  he  receive  for  his  flour  ? 

354.    To   find   the  Per  Cent,  the  Cost  and  the  Profit  of 
Loss  being  given. 

Divide  the  profit  or  loss  ly  the  cost.    (Prob.  Ill,  Per  ct.) 

NOTE. — When  the  cost  and  selling  price  are  given,  first  find  the 
profit  or  loss,  then  the^er  cent.    (Art.  339.) 

20.  If  I  buy  an  acre  of  land  for  $320,  and  sell  it  for  $80 
more  than  it  cost  me,  what  is  the  per  cent  profit  ? 

SOLUTION. — Per  cent=$8o  (gain)H-$320  (cost)=.25  or  25$.  Ans. 

21.  A  jockey  paid  $875  for  a  fast  horse,  and  sold  it  so 
as  to  lose  $250 :  what  per  cent  was  his  loss  ? 

22.  If  I  pay  22\  cts.  a  pound  for  lard,  and  sell  it  at  2\ 
cts.  advance,  what  per  cent  is  the  profit  ? 

23.  If  a  newsboy  pays  2\  cts.  for  papers,  and  sells  them 
at  \\  cent  advance,  what  per  npnt  is  his  profit? 

24.  If  a  speculator  buys  apples  at  $2.12^  a  barrel,  an$ 
Bells  them  at  $2.87^,  what  is  his  per  cent  profit? 

ANALYSIS. — $2.87^  —  82.12^=8.75   profit  per  bl.     Therefore,  $.75 
(gain)-5-$2.i2^  (co8t)=.35,5r  or  35/V/6-     (Art.  339,  n.) 

25.  If  I  sell  an  article  at  double  the  cost,  what  per  cent 
is  my  gain  ? 

26.  If  I  sell  an  article  at  half  the  cost,  what  per  cent  is 
my  loss  ? 

27.  If  I  buy  hats  at  $3,  and  sell  at  $5,  what  is  the  per 
cent  profit  ? 

28.  If  I  buy  hats  at  $5,  and  sell  at  $3,  what  is  the  per 
cent  loss  ? 

354.  How  find  the  per  cent  profit  or  loss,  when  the  cost  and  profit  or  loes  are 


250  PEOFIT     AND     LOSS. 

29.  If  a  man's  debts  are  $3560,  and  he  pays  only  $1780, 
what  per  cent  is  the  loss  of  his  creditors  ? 

30.  If  |  of  an  article  be  sold  for  £  its  cost,  what  is  the 
per  cent  loss  ? 

ANALYSIS.  —  If  J  are  sold  for  £  its  cost,  $  must  be  sold  for  ^  of  ^, 
or  £  the  cost,  and  -J  for  £  or  f  the  cost.  Hence,  the  loss  is  ^  the 
cost  •,  and  ^-*-$=.33^  or 


31.  If  you  sell  \  of  an  article  for  |  the  cost  of  the  whole, 
what  is  the  gain  per  cent  ? 

32.  If  I  sell  |  of  a  barrel  of  flour  for  the  cost  of  a  barrel, 
what  is  the  per  cent  profit  ? 

33.  If  a  milkman  sells  3  quarts  of  milk  for  the  price  he 
pays  for  a  gallon,  what  per  cent  does  he  make  ? 

34.  Bought  3  hhd.  of  molasses  at  85  cts.  per  gallon,  and 
sold  one  hhd.  at  75  cts.,  the  other  two  at  $i  a  gallon- 
required  the  whole  profit  and  the  per  cent  profit? 

) 
355.   To  find  the  Cost,  the  Profit  or  Loss  and  the  Per  Cent 

Profit  or  Loss  being  given. 

Divide  the  profit  or  loss  ~by  the  given  rate.  (Problem  IV, 
Percentage.) 

NOTE.  —  When  the  per  cent  profit  or  loss  is  an  aliquot  part  of 
100,  the  operation  may  often  be  abbreviated  by  using  this  aliquot 
part  as  the  divisor.  (Art.  341,  n.) 

35.  A  grocer  sold  a  chest  of  tea  at  25%  profit,  by  which 
he  made  $22^:  what  was  the  cost  ? 

SOLUTION.  —  Cost=$22.  50  (profit)-*-.  25  (rate)=$9o.  Ans. 
Or,  cost=  $22.50-:-^=  $90.  Ans.    (Art.  340,  n.) 

36.  A  speculator  lost  $1950  on  a  lot  of  flour,  which  was 
20%  of  the  cost:  required  the  cost? 

37.  Lost  65  cents  a  yard  on  cloths,  which  was  IT>%  °f 
the  cost:  required  the  cost  and  selling  price  ? 

ANALYSIS.  —  The  cost  =65  cts.  -H.I  3=  $5  ;  and  $5  —  $.6s=$4.35  the 
selling  price.  (Art.  340,  n.) 

355.  How  find  the  cost,  when  the  profit  or  loss  and  the  per  cent  profit  or  loss 
are  given, 


PEOFIT     AND     LOSS.  251 

38.  Gained  $2^  per  barrel  on  a  cargo  of  flour,  which 
was  2^% :  required  the  cost  and  selling  price  per  barrel  ? 

AKALYSIS. — 20^=4,  and   $2.50-7-^=112.50  cost,  and    $12.50+ 
j>2.5o=$i5,  selling  price. 

39.  If  I  sell  coffee  at  \o%  profit  I  make  10  cts.  a  pound: 
what  was  the  cost  ? 

40.  A  man  sold  a  house  at  a  profit  of  33$$,  and  thereby 
gained  $7500:  required  the  cost,  and  selling  price? 

41.  If  I  make  20%  profit  on  goods,  what  sum  must  I  lay 
out  to  clear  $3500,  and  what  will  my  sales  amount  to  ? 

42.  A  and  B  each  gained  $1500,  which  was  12^%  of  A's 
and  1 6%  of  B's  stock :  what  was  the  investment  of  each  ? 

43.  If  a  merchant  sells  goods  at  \Q%  profit,  what  must 
be  the  amount  of  his  sales  to  clear  $25000  ? 

44.  A  market  man  makes  \  a  cent  on  every  egg  he 
sells,  which  is  25%  profit:  what  do  they  cost  him,  and 
how  sell  them  ? 

356.  To  find  the  Cost,  the  Selling  Price  and  the   Per  Cent 
Profit  or  Loss  being  given. 

Divide  the  selling  price  by  i  plus  or  minus  the  rate  of 
profit  or  loss,  as  the  case  may  require.    (Prob.  V,  Per  ct.) 

45.  A  jockey  sold  two  horses  for  $168.75  eac^  '•>  on  one  ne 
made  1 2-^,  on  the  other  lost  1 2\% :  what  did  each  horse 
cost  him  ? 

SOLUTION. — Cost  of  one=$i68.75  (sel.  pr.)-=-i.i25  (i  +  rate)=$i5o. 
Cost  of  others $168.75  (sel-  pr.)-*-.875  (i— rate)=$iq2.86.  Ans. 

46.  By  selling  568  bis.  of  beef  at  $15^  a  barrel,  a  grocer 
lost  i2-|%:  what  was  the  cost?  Ans.  $10061. 71^. 

47.  Sold  5000  acres  of  land  at  $3^  an  acre,  and  thereby 
gained  22$:  what  was  the  cost? 

48.  Sold  a  case  of  linens  for  £27,  ios.,  making  a  profit 
of  25$:  what  was  the  cost? 


3t6.  How  find  the  cost,  when  the  sailing  price  and  the  per  cent  profit  »re  given  t 


252  PROFIT     AND     LOSS. 

49.  If  a  newsboy  sells  papers  at  4  cts.  apiece,  he  makes 
ZZ\%  '•  what  do  they  cost  him  ? 

357.    To  Mark  Goods  so  that  a  given   Per  Cent  may  be 
deducted,  and  yet  make  a  given  Per  Cent  profit  or  loss. 

50.  Bought  shoes,  at  $2.55  a  pair:  at  what  price  must 
they  be  marked  that  1 5%  may  be  deducted,  and  yet  be 
gold  at  20%  profit  ? 

ANALYSIS. — The  selling  price  is  120$  of  $2.55  (the  cost) ;  $2.55  x 
i.20=$3.o6,  the  price  at  which  they  are  to  be  sold.  »But  the  marked 
price  is  loofo  of  itself;  and  100^  —  15^=85^.  The  question  novr 
is,  $3.06  are  85$  of  what  sum  I  If  $3.06=-,^,  Tiu=$3-o6H-85,  or 
$.036;  and  -H}X=$3.6o.  Am.  (Art.  340.)  Hence,  the 

KULE. — Find  the  selling  price,  and  divide  it  by  i  minus 
the  given  per  cent  to  be  deducted ;  the  quotient  will  be  the 
marked  price. 

51.  Paid  56  cts.  apiece  for  arithmetics:  what  must  they 
be  marked  in  order  to  abate  5$,  and  yet  make  25%  ?% 

52.  If  mantillas  cost  $24  apiece,  at  what  price  must  they 
be  marked  that,  deducting  8$,  the  merchant  may  realize 
ZZ\%  profit? 

53.  When  apples  cost  $3.60  a  barrel,  what  must  be  the 
asking  price  that,  if  an  abatement  of  12-^  is  made,  there 
mil  still  be  a  profit  of  i6|^? 

54.  A  merchant  paid  87.^  cts.  a  yard  for  a  case  of  linen, 
which  proved  to  be  slightly  damaged :  how  must  he  mark 
it  that  lie  may  fall  25^,  and  yet  sell  at  cost? 

55.  A  goldsmith  bought  a  case  of  watches  at  $60:  how 
must  he  mark  them  that,  abating  4^',  he  may  make  20',  ? 

56.  If  cloths  cost  a  tailor  $4.50  a  yard,  at  what  price 
must  he  mark  them,  that  deducting  \Q%  he  will  make 
15  per  cent  profit? 


357.  When  the  selling  price  and  the  per  cent  profit  or  loss  are  given,  how  find 
the  per  cent  profit  or  loss  at  any  proposed  price  ?  How  find  what  to  mark  goods. 
that  a  given  per  cent  may  be  deducted,  and  yet  a  given  per  cent  profit  or  loss  be 


PBOFIT    AND    LOSS.      ^  253 

QUESTIONS     FOR     REVIEW.  <\'j 

1.  A  grocer  paid  23  cts.  a  Ib.  for  a  tub  of  butter  weighing  ^ 
65  Ibs.,  and  sold  it  at  18%  profit:  how  much  did  he  make?^ 

2.  Bought  a  house  for  $3865,  and  paid  $1583.62  for 
pairs:  how  much  must  I  sell  it  for  to  mate  i6f%?C 

3.  Paid  $2847  for  a  case  of  shawls,  and  $956  for  a  case 
of  ginghams;  sold  the  former  at  22%%  advance,  and  the 
latter  at  i  \%  loss :  what  was  received  for  both  tCil/l    -  ^ 

4.  A  jockey  bought  a  horse  for  $125,  Avhich  he  traded 
for  another,  receiving  $37  to. boot;  he  sold  the  latter  at 
25%  less  than  it  cost:  what  sum  did  he  lose?  G^ 

5.  What  will  750    shares  of  bank  stock  cost,  if  the  bro- 
kerage is  \%  and  the  stock  3%  above  par  ?  o  i  "V  l/l 

6.  Sold  3  tons  of  iron,  at  15^  cts.  a  pound,  and  charged      .* 
•$\%  commission :  what  were  the  net  proceeds ' .  • .  4U 

7.  Bought  wood 'at  $4-J  a  cord,  and  sold  it  for  1 6 1 :  re- 
quired the  per  cent  profit  ?—     •  1 

•~"S7  A  shopkeeper  buys  thread  at  4  cts.  a  spool,  aivd  sells 
at  6%  cts.:  Avhat  per  cent  is  his  profit  and  how  much 
would  he  make  on  1000  gross?  C>  "*  0  •• 

9.  Bought  1650  tons  of  ice,  at  $12 ;  one  half  melted,  and 
sold  the  rest  at  $i  a  hund. :  what  per  cent  was  the  loss?Of- .:, 

10.  The  gross  proceeds  of  a  consignment  of  apples  was 
$1863.75,  and  the  agent  deducted  $96.9 1£  for  selling:  re- 
quired the  per  cent  commission  and  the  net  proceeds? 

11.  A  lady  sold  her  piano  for  f  of  its  cost:  what  per 
cent  was  the  loss  ? 

12.  If  I  pay  $2  for  3  Ibs.  of  tea,  and  sell  2  Ibs.  for  $3, 
what  is  the  per  cent  profit? 

13.  A  man  sold  his  house  at  20$  above  cost,  and  therely 
made  $1860:    required  the  cost  and  the  selling  price.? 

14.  A  miller  sells  flour  at   15$  more  than  cost,  and 
makes  $1.05  a  bar.:  what  is  the  cost  and  selling  price  ? 

15.  Lost  25  cts.  a  pound  on  indigo,  which  was  \2\%  of 
the  cost :  required  the  cost  and  the  selling  price. 


254  PKOFIT     AND     LOSS. 

1 6.  A  commission  merchant  received  $260  for  selling 
a  quantity  of  provisions,  which  was   5%:  required  the 
amount  of  sales  and  the  net  proceeds. 
/- 1 7.  A  broker  who  charges  \%,  received  $60  for  selling  a 
quantity  of  uncurrent  money  :  how  much  did  he  sell  ? 

18.  A  grocer  makes  $1.25  a  pound  on  nutmegs,  which 
is  100%  profit :  what  does  he  pay  for  them  ? 

19.  A  merchant  sold  a  bill  of  white  goods  for  $7000, 
and  made  zzk%  '•  required  the  cost  and  the  sum  gained. 

20.  Sold  a  hogshead  of  oil  at  93^  cts.  a  gallon,  and 
made  i8f$:  required  the  cost  and  the  profit. 

21.  A  man,  by  selling  flour  at  $12^  a  barrel,  makes  25%: 
what  does  the  flour  cost  him  ? 

22.  Made  12^  on  dry  goods,  and  the  amount  of  sales 
was  $57725  :  required  the  cost  and  the  sum  made. 

23.  Sold  a  quantity  of  metals,  and  retaining  $\%  com- 
mission, sent  the  consignor  $15246:  required  the  amount 
of  the  sale  and  the  commission. 

24.  Sold  250  tons  of  coal  at  $6|,  and  made  12^' :  what 
per  cent  would  have  been  my  profit  had  I  sold  it  for 
$8  a  ton  ? 

25.  A  merchant  sold  out  for  $18560,  and  made  15%  on 
his  goods :  what  per  cent  would  he  have  gained  or  lost  by 
selling  for  $15225  ? 

26.  Paid  $40  apiece  for  stoves :  what  must  I  ask  that  I 
may  take  oif  20%,  and  yet  make  20$  on  the  cost  ? 

27.  If  a  bookseller  marks  his  goods  at  25$  above  cost, 
and  then  abates  25%,  what  per  cent  does  he  make  or  lose-3 

28.  A  grocer  sells  sugar  at  2\  cts.  a  pound  more  than 
oost,  and  makes  20%  pro6t :  required  the  selling  price. 

29.  A  man  bought  2500  bu.  wheat,  at  3>if ;  3200  bu.  corn, 
at  87^  cts. ;  4000  bu.  oats,  at  25  cts.;  and  paid  $450  freight: 
he  sold  the  wheat  at  5$  profit,  the  corn  at  11%  loss,  and 
the  oats  at  cost ;  the  commission  on  sales  was  5% :  what 
was  his  per  cent  profit  or  loss  ? 


.».  29- 


INTEREST. 

358.  Interest  is  a  compensation  for  the  use  of  money. 
It  is  computed  at  a  certain  per  cent,  per  annum,  and  em- 
braces five  elements  or  parts:  the  Principal,  the  Rate,  the 
Interest,  the  Time,  and  the  Amount. 

359.  The  Principal  is  the  money  lent. 
The  JZate  is  the  per  cent  per  annum. 
The  Interest  is  the  percentage. 

The  Time  is  the  period  for  which  the  principal  draws 
interest 
The  Amount  is  the  sum  of  the  principal  and  interest. 

NOTES. — i.  The  term  per  annum,  from  the  Latin  per  and  annus, 
signifies  by  the  year. 

2.  Interest  differs  from  the  preceding  applications  of  Percentage 
only  by  introducing  time  as  an  element  iu  connection  with  per  cent. 
The  terms  rate  and  rate  per  cent  always  mean  a  certain  number  of 
hundredths  yearly,  and  pro  rata  for  longer  or  shorter  periods. 

360.  Interest  is  distinguished  as  Simple  and  Compound. 
Simple  Interest  is   that  which   arises  from  the 

principal  only. 

Compound  Interest  is  that  which  arises  both  from 
the  principal  and  the  interest  itself,  after  it  becomes  due. 

361.  The  Legal  Kate  of  interest  is  the  rate  allowed 
by  law.    Rates  higher  than  the  legal  rate,  are  called  usury. 

NOTES. — i.  In  Louisiana  the  legal  rate  is     -        -               -  $%. 
In  the  N.  E.  States,  N.  C.,  Penn.,  Del.,  Md.,  Va.,  W.  Va., 
Tenn.,  Ky.,  0.,  Mo.,  Miss.,  Ark.,  Flor,  la.,  111.,  Ind.,  the 

Dist.  of  Columbia,  and  debts  due  the  United  States,     -  b%. 

In  N.  Y .,  N.  J.,  S.  C.,  Ga.,  Mich.,  Min.,  and  Wis.,        -        -  1%, 

In  Alabama  and  Texas, 8$, 

In  Col ,  Kan.,  Neb.,  Nev.,  Or.,  Cal.  and  Washington  Ter.,  -  io$>. 

l.  In  some  States  the  law  allows  higher  rates  by  special  agreement. 

3.  When  no  rate  is  specified,  it  is  understood  to  be  the  It  gal  rate 

358.  What  is  interest  ?    359.  The  principal  ?    The  rate  ?    The  int.  T    The  time  I 
Fbe  aint.  ?    360.  What  is  simple  interest  ?    Compound  ? 


256  INTEREST. 

THE    SIX    PER    CENT    METHOD. 

362.  Since  the  interest  of  $i  at  6  per  cent  for  1 2  months 
or  i  year,  is  6  cents,  for  i  month  it  is  i-tivelfth  of  6  cents 
or  \  cent ;  for  z  months  it  is  2  halves  or  i  cent;  for  3  mos. 
i^  cent,  for  4  mos.  2  cents,  etc.    That  is, 

The   interest  of  ^i,  at  6  per  cent,  for  any  number  of 
months,  is  half  as  many  cents  as  months. 

363.  Since  the  interest  of  $i  at  6  per  cent  for  30  days 
or  i  mo.,  is  5  mills  or  \  cent,  for  i  day  it  is  -^  of  5  mills 
or  £  mill ;  for  6  days  it  is  6  times  £  or  i  mill ;  for  1 2  days, 
2  mills  ;  for  15  days,  z\  mills,  etc.     That  is, 

The  interest  of  81,  at  6  per  cent,  for  any  number  of  days, 
is  i-sixth  as  many  mills  as  days.    Hence, 

364.  To  find  the  Interest  of  $1,  at  6  pep  cent,  for  any 
given  time. 

Take  half  the  number  of  months  for  cents,  and  one  sixth  of 
the  days  for  mills.  Their  sum  will  be  the  interest  required. 
NOTES. — i.  When  the  rate  is  greater  or  less  than  6  per  cent,  the 
interest  of  $i  for  the  time  is  equal  to  the  interest  at  6  per  cent 
increased  or  diminisJicd  by  such  a  part  of  itself  as  the  given  rate 
exceeds  or  falls  short  of  6  per  cent.  Thus,  if  the  rate  is  7%,  add  ^  to 
the  int.  at  6;?  ;  if  the  rate  is  5$,  subtract  £,  etc. 

2.  In  finding  i-sixth  of  the  days,  it  is  sufficient  for  ordinary  pur. 
poses  to  carry  the  quotient  to  tenths  or  hundredths  of  a  mill. 

3.  When  entire  accuracy  is  required,  the  remainder  should  be 
placed  over  the  6,  and  annexed  to  the  multiplier. 

1.  Int.  of  $i,  at  1%  for  9  m.  18  d.  ?  Ans.  $.056. 

2.  Int.  of  $i,  at  5$  for  n  m.  21  d.  ?       AH*.  8.04875. 

3.  What  is  the  int.  of  $i,  at  6%  for  7  m.  3  d.  ? 

4.  What  is  the  int.  of  $i,  at  6%  for  n  m.  13  d.  ? 

Note.  Wherein  does  interest  differ  from  the  preceding  applications  of  per- 
centage? 361.  The  legal  rate?  What  is  usury  ?  362.  What  is  the  interest  of  $i 
sit  6  per  cent  for  months?  363.  For  daye?  364.  How  find  the  int.  of  $i  for  any 
[riven  time  ?  Note.  When  the  rate  is  greater  or  less  thau  6  per  cent,  to  what  is 
the  interest  of  $i  eqnal  ? 


INTEREST.  257 

REM.  —  The  relation  between  the  principal,  the  interest,  the  rate, 
the  time,  and  the  amount,  is  such,  that  when  any  three  of  them  are 
given,  the  others  can  be  found.  The  most  important  of  these 
problems  are  the  following  : 

PROBLEM    I. 

365.    To  find   the   Interest,  the   Principal,  the   Rate,  and 
Time  being  given. 

1.  What  is  the  interest  of  $150.25  for  i  year  3  months 
and  1  8  days,  at  6%? 

ANALYSIS.  —  The  interest  of  $i  for  15  m.=.o75  OPERATION. 

iSd.  =.003  Prin.  $150.25 

"             i  y.  3  m.  18  d.  =To7lT  .078 

Now  as  the  int.  of  $i  for  the  given  time  and  120200 

rate  is  $.078  or  .078  times  the  principal,  the  int.  105175 

of  $150.25   must   be  .078  times  that  sum;  and  $ii.7iqco 
$150.25  x  .078  =  $!!.  71950.     Hence,  the 

RULE.  —  Multiply  the  principal  by  the  interest  of  $i  for 
the  time,  expressed  decimally.  (Art.  336.) 

For  the  amount,  add  the  interest  to  the  principal, 

REM.  —  i.  The  amount  may  also  be  found  by  multiplying  the 
principal  by  i  plus  the  interest  of  $i  for  the  time.  (Art.  337.) 

2.  When  the  rate  is  greater  or  less  than  6%,  it  is  generally  best  to 
find  the  interest  of  the  principal  at  6%  for  the  given  time  ;  then  add 
to  or  subtract  from  it  such  a  part  of  itself,  as  the  given  rate  exceeds 
or  falls  short  of  6  per  cent.     (Art.  364.) 

3.  In  finding  the  time,  first  determine  the  number  of  entire  calendar 
months  ;  then  the  number  of  days  left. 

4.  In  computing  interest,  if  the  mills  are  5  or  more,  it  is  customary 
to  add  i  to  the  cents  ;  if  less  than  5,  they  are  disregarded. 

Only  three  decimals  are  retained  in  the  following  Answers,  and 
each  is  found  by  the  rule  under  which  the  Ex.  is  placed. 

2.  What  is  the  amt.  of  $150.60  for  i  y.  5  ra.  15  d.at  6%  .' 
ANALYSIS.  —  Int.  of  Si  for  i  y.  5  mos.  or  17  mos.,  equals  .085  ;  15  d. 

it  equals  .0025  ;  and  .085  +  .0025  =  .  0875  the  multiplier.     Now  $150.60 
x.o875=$i3.  1775,  int.    Finally,  $i5O.6o  +  $i3.i775=$i63.  7775,  Ans. 


^5.  How  find  the  interest,  when  the  principal,  rate,  and  time  are  given? 
How  find  the  amount,  when  the  principal  and  interest  are  given  f  Hem,  How 
else  is  the  amount  found  ': 


258  INTEBEST. 

3.  Find  the  interest  of  $31.75  for  r  yr.  4  mos.  at  6fc. 

4.  What  is  the  ink  of  $49.30  for  6  mos.  24  d.  at  6%  ? 

5.  What  is  the  int.  of  $51.19  for  4  mos.  3  d.  at  yfa  ? 

6.  What  is  the  int.  of  $142.83  for  7  mos.  18  d.  at  5%?  . 

7.  What  is  the  int.  of  $741.13  for  n  mos.  21  d.  at  6$? 

8.  What  is  the  int.  of  $968.84  for  i  yr.  lomos.  26d.-at6$? 

9.  What  is  the  int.  of  $639  for  8  mos.  29  d.  at  7%? 
co.  What  is  the  int.  of  $741:13  for  7  mos.  17  d.  at 

11.  What  is  the  int.  of  $1237.63  for  3  mos.  3<i.  at  8 

12.  What  is  the  int.  of  $2046^  for  13  mos.  25  d.  at  4 

13.  What  is  the  int.  of  $3256.07  for  i  rn.  and  3  d.  at 

14.  Find  the  amount  of  $630.37^  for  9  mos.  15  d.  at  i 

15.  Find  the  amount  of  $75.45  for  13  mos.  19  d.  at 
T  6.  Find  the  amount  of  $2831.20  for  2  mos.  3  d.  at 

17.  Find  the  amount  of  $356.81  for  3  m.  n  d.  at 

18.  Find  the  amount  of  $2700  for  4  mos.  3  d.  at 

19.  Eequired  the  amount  of  $5000  for  33  days  at  1 

20.  Eequired  the  int.  of  $12720  for  2  mos.  17  d.  at 

21.  What  is  the  amt.  of  $221.42  for  4  mos.  23  d.  at 

22.  What  is  the  int.  of  $563. 16  for  4  mos.  at  2%  a  month  ? 
SUGGESTION. — At  2  %  a  month,  the  int.  of  $i  for  4  mos.  is  $.08. 

23.  What  is  the  int.  of  $7216.31  for  3  mos.  at  \%  a  month? 

24.  Find  the  int.  of  $9864  for  2  mos.  at  2\%  a  month  ? 

25.  Find  the  amt.  of  $3540  for  17  mos.  10  d.  at  l\%t 

26.  What  is  the  interest  on  $650  from  April  i7th,  1870, 
to  Feb.  8th,  1871,  at  6^? 

ANALYSIS. — 1871  y.  2  m.  8  d.    \  Int.  $i  for  9  m.=.o45  $650 

1870  y.  4  m.  17  d.  v    "        "    21  d.  =.0035  .0485 


Time,        o  y.  g  m.  21  d.  \          Multiplier,      .0485       $31-525 

27.  What  is  the  interest  on  $1145  from  July  4th,  1867, 
to  Oct.  3d,  1868,  at  -]%. 

28.  What  is  the  interest  on  a  note  of  $568.45  from  May 
2ist,  1861,  to  March  25th,  1862,  at  5$? 

29.  Eequired  the  amount  of  $2576.81  from  Jan. 
1871,  to  Dec.  1 8th,  1871,  at  ^%. 


INTEREST.  259 


METHOD    BY    ALIQUOT    PARTS. 

366.    To  find   the    Interest   by  Aliquot   Parts,  the   Principal, 
Rate,  and  Time  being  given. 

i.  What  is  the  interest  of  $137  for  3  y.  i  m.  6  d.  at  5^? 

ANALYSIS. — As  fhe  rate  is  5  % ,  the  interest  for  $  1 3  7  prill . 

i  year  is  .05  (-[fh;)  of  the  principal,     ...  .05  rate. 

Now  $137 x. 05 =$6.85, 1 2)16.85  int.  i  y. 

Again,  the  int.  for  3  y.  is  3  times  as  much  as  for  i  y.,  3  .7- 

And  $6.85  (int.  i  y.)  x  3=120.55,        -        -        -  20.55  int.  3  y. 
The  int.  for  i  m.  is  -fa  of  int.  for  i  yr. ;  and 

$6.85-t-i2=.57,  5)        -57   "   im. 
The  int.  for  6  d.   is   i   of  int.   for   i   m.;  and 

$-57-»-5=-«4.  -"4  "  6  d. 

Adding  these  partial  interests  together,  we  have  $21.234,  Ans. 

the  answer  required.     Hence,  the 

KULE. — For  one  year. — Multiply  the  principal  by  the 
rate,  expressed  decimally. 

For  two  or  more  years. — Multiply  the  interest  for  i  year 
by  the  number  of  years. 

For  months. — Take  the  aliquot  part  of  i  year's  interest. 

For  days. — Take  the  aliquot  part  of  i  month's  interest. 

That  is,  for  i  m.,  take  TTj  of  the  int.  for  i  y. ;  for 
i  m.,  £ ;  for  3  m.,  J,  etc. 

For  i  d.,  take  ^  of  the  int.  for  i  m. ;  for  2  d.,  Tx^ ;  for 
6  d.,  £;  for  10  d.,  £,  etc. 

NOTE. — This  method  has  the  advantage  of  directness  for  different 
rates ;  but  in.  practice,  the  preceding  method  is  generally  shorter 
and  more  expeditious. 

2.  What  is  the  int.  of  $143.21  for  2  y.  5  m.  8  d.  at 

3.  What  is  the  int.  of  $76.10  for  i  y.  3  m.  5  d.  at 


366.  How  find  the  interest  for  i  year,  at  any  given  rate,  by  aliquot  parts? 
How  for  2  or  more  years  ?  For  months  ?  For  days  ?  What  part  for  i  month  * 
Cor  3  mos.?  For  6  mos.?  For  i  day  ?  For  5  d.?  Foriod,?  For  15 d.?  Forzod.T 


260  INTEREST. 

4.  What  is  the  int.  of  $95.31  for  8  m.  20  d.  at 

5.  What  is  the  int.  of  $110.43  frr  l  Jr-  6  m-  I0 

6.  What  is  the  int.  of  $258  for  3  yrs.  7  m  at  S 

7.  What  is  the  interest  of  $205.38  for  5  yrs.  at 

8.  Find  the  interest  of  $361.17  for  n  months  at  &%? 
•9.  Find  the  interest  of  $416.84  for  19  days  at  7$$? 

10.  At  i\%,  what  is  the  int.  of  $385.20  for  i  yr.  13  d. 

1 1.  At  5^%,  what  is  the  int.  of  $1000  for"  i  y.  i  m.  3  d.  ? 

12.  At  Sfc,  what  is  the  int.  of  $1525.75  for  3  months? 

13.  Kequired  the  interest  of  $12254  for  2\  years  at  8%? 

14.  What  is  the  amount  of  $20165  f°r  5  m-  J7  d.  at  7$? 

METHOD    BY    DAYS. 

367.    To  find  the  Interest  by  Days,  the  Principal,  Rate, 
and  Time  being  given. 

i.  What  is  the  interest  of  $350  for  78  days  at  6%? 

ANALYSIS. — The  interest  of  $i  at  6$  for  i  day  $350 

is  £  of  a  mill  or  5(f,,r,  dollar.     (Art.  363.)     There-  78 

fore  for  78  days  it  must  be  78  times  gBLno  =  Bo»Tf  ^2800 

dol.,  or  sJfiu  times  the  principal.     Again,  since 
the  int.  of  $i  for  78  days  is  ^80  times  the  princi- 


pal, the  interest  of  $350  for  the  same  time  and      6{ooo)$27j3OO 
rate  must  be  sftijij  times  that  sum.     In  the  opera-         Ans.  $4.55 
tion,  we  multiply  the  principal  by  78,  the  numera- 
tor, and  divide  by  the  denominator  6000.     Hence,  the 

EULE. — Multiply  the  principal  by  the  number  of  days,  and 
divide  by  6000.  TJie  quotient  will  be  the  interest  at  6%. 

For  other  rates,  add  to  or  subtract  from  the  interest  at 
6  per  cent,  such  a  part  of  itself  as  the  required  rate  is 
greater  or  less  than  6fc.  (Art.  364.) 

NOTE. — This  rule,  though  not  strictly  accurate,  is  generally  used 
by  private  bankers  and  money-dealers.  It  is  based  upon  the  suppo- 
sition that  360  days  are  a  year,  which  is  an  error  of  five  days  or  Va 
of  a  year.  Hence,  the  result  is  -fa  too  large.  When  entire  accuracy 
is  required,  the  result  must  be  diminished  by  -fa  part  of  itself. 

367.  How  compute  interest  by  days  ?    Note.  Upon  what  is  this  method  founded  T 


INTEBEST.  261 


2.  A  note  for  $720  was  dated  April  1 7th,  1871 :  what  was 
the  interest  on  it  the  i6th  of  the  following  July  at  1%  ? 
Omitting  the  day  of  the  first  date.   (Art.  339,  n.) .         $720  prin. 
April  has  30  days— 17  d.= 13  d.  90  days. 

May     "  31  d.        6|ooo)64|8oo 

June    "  30  d.  6)  1 0.80  int.  6%. 

July     "  i6d.  i. 80    «    1%. 

Time    "  =90  d.         Ans.  $12.60  int.  1%. 

3.  What  is  the  interest  of  $517  for  33  days  at  6%? 

4.  What  is  the  interest  of  $208.75  f°r  63  days  at  6%? 

5.  What  is  the  interest  of  $631.15  for  93  days  at  7$? 

6.  Find  the  interest  of  $1000  for  100  days  at  5%? 

7".  Find  the  amount  of  $1260.13  for  120  days  at  6%? 

8.  Required  the  interest  of  $3568.17  for  20  days  at 

9.  Required  the  amount  of  $4360.50  for  3  days  at  ,  ,_ 

10.  Find  the  interest  of  $5000  from  May  2ist  to  the  /  $  /d* 
5th  of  Oct.  following  at  6%  ? 

n.  Find  the  interest  of  $6523  from  Aug.  i2th  to  the 
5th  of  Jan.  following  at  7%? 

12.  Find  the  interest  of  $7510  from  Jan.  5th  to  the 
loth  of  the  following  March,  being  leap  year,  at  6%  ? 

PROBLEM    II. 

368.  To  find  the  Rate,  the  Principal,  the  Interest,  and  the 
Time  being  given. 

i.  A  man  lent  his  neighbor  $360  for  2  yrs.  3  mos.,  and 
received  $48.60  interest :  what  was  the  rate  of  interest  ? 

ANALYSIS.— The  int.  of  $360  for     $360  x  .01  =$3.60  int.  i    y. 
i  year  at  i  %  is  $3.60;  and  for  2 }  y.       3.60  x  2\  =  $8.10    "    2\  y. 
$8.10.     Now,  if  $8.10  is  \%  of  the          $8.io)$48.6o 
principal,  $48  60  must  be  as  many  luTTT'-npr  of 

,  -ii  /fco.    \J   pel  \>u. 

per    cent   as    $8.10  are  contained 

times  in  $48.60,  which  is  6.    Therefore  the  rate  was  b%.     Hence,  the 

RULE. — Divide  the  given  interest  by  the  interest  of  the 
principal  for  the  tim,e,  at  i  per  cent. 

368.  How  find  the  rate,  when  the  principal,  rate,  and  time  are  given  f 


262  INTEREST. 

NOTE. — Sometimes  the  amount  is  mentioned  instead  of  the  prin. 
cipal,  or  the  intercut.  In  either  case,  the  principal  and  interest  may 
be  said  to  be  given.  For,  the  amt.=the  prin.  +  int. ;  hence,  amt.— 
int.=the  prin. ;  and  amt.— prin.=the  int.  (Arts.  365,  101,  Def.  17.) 

2.  If  $600  yield  $10.50  interest  in  3  months,  what  is  the 
rate  per  cent  ? 

3.  At  what  rate  will  $1500  pay   me  $52.50  interest 
semi-annually  ? 

4.  At  what  rate  will  $1000  amount  to  $1200  in  3  y.  4  m.? 

5.  A  lad  at  the  age  of  14  received  a  legacy  of  $5000, 
which  at  21  amounted  to  $7800 :  what  was  the  rate  of  int.  ? 

6.  A  man  paid  $9600  for  a  house,  and  rented  it  for  $870 
a  year:  what  rate  of  interest  did  he  receive  for  his  money  ? 

7.  At  what  rate  of  int.  will  $500  double  itself  in  12  years  ? 

8.  At  what  rate  will  $1000  double  itself  in  20  y.  ?  In  10  y.  ? 

9.  At  what  rate  will  $1250  double  itself  in  14!  years? 
10.  At  what  rate  must  $3000  be  put  to  double  itself  in 

1 6f  years  ? 

PROBLEM    III. 

369.  To  find  the  Time,  the   Principal,  the  Interest,  and  the 
Rate  being  given. 

i.  Loaned  a  friend  $250  at  6%,  and  received  $45  interest: 
how  long  did  he  have  the  money  ? 

ANALYSIS. — The  int.  of  $250  at  $250  x  .o6  =  $i5  int.  for  i  y. 
6%  for  i  yr.  is  $15.  Now  if  $15  $I5)$45  int. 

int    require    the  given  principal  ^wsTT^years. 

i  yr.  at  6#,  to  earn  $45  int.,  the 

same  principal  will  be  required  as  many  years  as  $15  are  contained 
times  in  $45;  and  $45-i-$i5=3.  He  therefore  had  the  money  3 
years.  Hence,  the 

RULE. — Divide  tlie  given  interest  by  the  interest  of  the 
principal  for  i  year,  at  the  given  rate. 

Note.  How  find  the  principal,  when  the  amount  and  interest  are  given  ?  How 
find  the  interest,  when  the  amount  and  principal  are  given  ?  369.  How  fad  thw 
time,  when  the  principal,  interest,  and  rate  are  given  ? 


INTEREST.  263 

NOTES.— i.  If  the  quotient  contains  decimals,  reduce  them  to 
months  and  days.  (Art.  294.) 

2.  If  the  amount  is  given  instead  of  the  principal  or  the  interest, 
find  the  part  omitted,  and  proceed  as  above.  (Art.  368.  n.) 

2.  In  what  time  will  $860  amount  to  $989  at  6%  ? 
ANALYSIS.— The  amount  $989— $86o=$i29.oo  the  int.,  and  the 

•nt.  of  $860  for  i  year  at  6fc  is  $51.60.     Now  $i29H-$5i.6o=:2.5,  or 
iV  years. 

3.  In  what  time  will  $1250  yield  $500,  at  7%  ? 

4.  In  what  time  will  $2200  yield  $100,  at  6%  ? 

5.  In  what  time  will  $10000  yield  $200,  at  8$? 

6.  In  what  time  will  $700  double  itself,  at  6%  ? 
SOLUTION.— The  int.  of  $700  for  i  year  at  6$,  is  $42  ;  and  $700-*- 

$42 =i6|  years.  Ans. 

7.  How  long  must  $i  200  be  loaned  at  7%  to  double  itself? 

8.  In  what  time  will  $7500  amount  to  $15000,  at  6$? 
9.  In  what  time  will  $10000  amount  to  $25000,  at  8$? 

PROBLEM    IV. 

370.  To  find  the  Principal,  the  Interest,  the  Rate,  and  the 
Time  being  given. 

i.  What  principal,  at  6%,  will  produce  $60  int.  in  2\  yrs.  ? 
ANALYSIS. — 2-J-  y.— 30  mos. ;  therefore  the 
int.  of  $i  for  the  given  time  at  6</c  is  15  cents.        2\  v.  =  3o  m. 
Now  as  $.15  is  the  int.  of  $i   for  the  given     \~±  ~p 4j— &  JK 
time  and  rate,  $60  must  be  the  int.  of  as  many 


dollars  as   $.15   are  contained   times  in   $60  ; 

and   $60  -H$.  1  5  =  $400,  the  principal  required.       Am.  $400  prin. 

Hence,  the 


.  —  Divide  the  ffiren  inferfist  by  the  interest  of  $*  for 
fie  ffiren  time  and  rate,  expressed  decimally. 

2.  At  6#,  what  principal  will  yield  $100  in  i  year? 

3.  At  7$,  what  principal  will  yield  $105  in  6  months? 

370.  How  find  the  principal,  when  the  interest,  rate,  and  time  are  given  ? 


£64  INTEEEST. 

4.  What  sum, at  5%,  will  gain  $175  in  i  year  6  months? 

5.  What  sum  must  be  invested  at  6%  to  pay  the  ground 
rent  of  a  house  which  is  $150  per  annum? 

6.  A  gentleman  wished  to  found  a  professorship  with 
annual  income  of  $2800:  what  sum  at  1%  will  produce  it? 

7.  A  man  invested  his  money  in  6%  Government  stocks, 
and  received  $300   semi-annually :   what  was  the    sum 
invested  ? 

8.  A  man  bequeathed  his  wife  $1500  a  year:  what  sum 
must  he  invest  in  6%  stocks,  to  produce  this  annuity  ? 

PROBLEM    V. 

371.  To  find  the  Principal,  the  Amount,  the  Rate,  and  the 
Time  being  given. 

1.  What  principal  will  amount  to  $508.20  in  3  y.  at  6%: 

ANALYSIS. — 3  y.=3&  m.  ;    therefore  the 

int.  of  $i  for  the  given  time  at  6%  is  18  cts.,  3  years=36  m. 

and  the  amt.=$i.i8.     (Art.  365.)    Now  as  jnt.  $I=    <$.X8 

§1.18  is  the  amt.  of  $i  principal  for  the  given  amt   $  I  =  l>  I  1 8 

time  and  rate,  $508.20  must  be  the  amount 

of  as  many  dollars  principal  as  $1.18  are  con-  *  *J    j^5°   -2O 

tained  times  in  $508.20  ;  and  $5o8.2O-r-$i.i8  Ans.  $430.68 

=$430.68,  the  principal  required.   Hence,  the 

EULE. — Divide  the  given  amount  by  the  amount  of  $i  for 
the  given  time  and  rate,  expressed  decimally. 

2.  What  principal,  at  6%,  will  amount  to  $250  in  i  year? 

3.  What  principal,  at    7$,  will    amount  to   $356   in 
i  year  3  months  ? 

4.  What  sum  must  a  father  invest  at  6£,  for  a  son  19 
years  old,  that  he  may  have  $10000  when  he  is  21  ? 

5.  What  sum  loaned  at  i%  a  month  will  amount  to 
$1000  in  i  year? 

6.  What  sum  loaned  at  2%  a  month  will  amount  to 
$6252  in  6  mos.  ? 

371.  How  find  the  principal,  when  the  amount,  rate,  and  time  are  pivu«  f 


PROMISSORY   iNOTES. 

372.  A  Promissory  Note  is  a  written  promise  to 
pay  a  certain  sum  at  a  specified  time,  or  on  demand. 

NOTE. — A  Note  should  always  contain  the  words  "  value  re- 
ceived ;  •'  otherwise,  the  holder  may  be  obliged  to  prove  it  was 
given  foi  a  consideration,  in  order  to  collect  it. 

373.  The  person  who  signs  a  note  is  called  the  maker  or  drawer* 
the  person  to  whom  it  is  made  payable,  the  payee;  and  the  person 
who  has  possession  of  it,  the  holder. 

374.  A.  Joint  Note  is  one  signed  by'two-or  more  persons. 

375.  The  Face  of  a  Note  is  the  sum  whose  payment  is 
promised.    This  sum  should  be  written  in  wards  in  the  body  of  thfa 
note,  and  in  figure*  at  the  top  or  bottom. 

376.  When  a  note  is  to  draw  interest  from  its  date,  it  should 
contain  the  words  "  with   interest ;"  otherwise  no  interest  can  be 
collected.     For  the  same  reason,  when  it  is  to  draw  interest  from  a 
particular  time  after  dtite,  that  fact  should  be  specified  in  the  note. 

All  notes  are  entitled  to  legal  interest  after  they  become  due, 
whether  they  draw  it  before,  or  not. 

377.  Promissory  Notes  are  of  two  kinds ;  negotiable,  and  non- 
negotiable. 

378.  A  Negotiable  Note  is  a  note  drawn  for  the  payment  of 
money  to  "  order  or  bearer,"  without  any  conditions. 

A  Non-Negotiable  Note  is  one  which  is  not  made  payable  to 
"  order  or  bearer,"  or  is  not  payable  in  money. 

NOTES. — i.  A  note  payable  to  A.  B.,  or  "  order,"  is  transferable  by 
indorsement;  if  to  A.  B,  or  "bearer,"  it  is  transferable  by  delivery. 
Treasury  notes  and  bank  bills  belong  to  this  class. 

2.  If  the  words  "  order"  and  "  bearer"  are  both  omitted,  the  note 
can  be  collected  only  by  the  party  named  in  it. 

379.  An  Indorser  is  a  person  who  writes  his  name  upon  the 
back  of  a  note,  as  security  for  its  payment. 

380.  The  jHaturit}/  of  a  Note  is  the  day  it  becomes  legally 
due.    In  most  of  the  United  States  a  note  does  not  become  legally  due 

372.  What  ia  a  promissory  note  ?  Whnt  T>nrtionlnr  wnrrls  phonlrt  n  note  rnn- 
Jaln  ?  373.  What  is  the  one  who  signs  it  called  ?  The  one  to  whom  it  is  payable  ? 
The  oce  who  has  it?  374.  A  joint  note?  375.  The  face  of  a  note?  377.  Of 
how  many  kinds  are  notes  ?  378.  A  negotiable  note  ?  A  non-negotiable  note  ? 
Kate.  How  is  the  former  transferable?  Why  is  the  latter  not  negotiable r 
3;  j.  What  is  an  indorscr  ?  380.  The  maturity  of  a  note  ? 

12 


266  INTEREST. 

until  three  days  after  the  time  specified.    These  three  days  are  called 
days  of  grace.    Hence,  a  note  matures  on  the  last  day  ot  grace. 

38X.  When  a  note  is  given  for  any  number  of  months,  calendar 
'months  are  always  to  be  understood. 

382.  If  a  note  is  payable  on  demand,  it  is  legally  due  as  soon  as 
presented.     If  no  time  i&  specified  for  the  payment,  it  is  understood 
to  be  on  demand. 

383.  A  Protest  is  a  written   declaration  made  by  a  notary 
public,  that  a  note  has  been  duly  presented  to  the  maker,  and  has 
not  been  paid. 

NOTE. — A  protest  must  be  made  out  the  day  the  note  or  draft 
matures,  and  sent  to  the  indorser  immediately,  to  hold  him  responsible. 

ANNUAL   INTEREST. 

384.  When  notes  are  made  "  with  interest  payable  annually," 
some  States  allow  simple  legal  interest  on  each  year's  interest  from 
the  time  it  becomes  due  to  the  time  of  final  settlement. 

385.  To  compute  Annual  Interest,  the  Principal,  Rate, 
and  Time  being  given. 

1.  "What  is  the  amount  due  on  a  note  of  8500,  at  6%,- 
in  3  years  with  interest  payable  annually  ? 

SOLUTION. — Principal,  §500.00 

Int.  for  i  y.  is  $30 :  for  3  y.  it  is  $30  x  3,  or  90.00 

Int.  on  ist  annual  int.  for  2  y.  is  3-6o 

2d       "        "      "    xy.  is  i. 80 

The  Amt.  is  $595 .40.    Hence,  the  Ans.  $595.40 

RULE. — Find  the  interest  on  the  principal  for  the  given 

time  and  rate;  also  find  the  simple  legal  iniersst  on  each 

year's  interest  for  the  time  it  has  remained  unpaid. 

Tlie  sum  of  the  principal  and  its  interest,  with  the 
interest  on  the  unpaid  interests,  will  be  the  amount. 

2.  What  is  the  amount  of  a  note  of  $1000  payable  in 
4  years,  with  interest  annually,  at  7$?      Ans.  $1309.40 

382.  Wlien  is  a  note  on  demand  due?    383.  What  ia  a  protest?    385.  How 
compute  annual  interest? 


PARTIAL   PAYMENTS. 

386.  Partial  Payments  are  parts  of  a  debt  paid 
at  different  times.  The  sums  paid,  with  the  date,  are 
usually  written  on  the  back  of  the  note  or  other  obliga- 
tion, and  are  thence  called  indorsements. 

UNITED    STATES    RULE. 

387.  To  compute  Interest  on  Notes  and  Bonds,  when  Partial 
Payments  have  been  made. 

1.  Find  the  amount  of  the  principal  to  the  time  of  the 
first  payment.    If  the  payment  equals  or  exceeds  the  interest, 
subtract  it  from  this  amount, and,  considering  the  remainder 
a  new  principal,  proceed  as  before. 

II.  If  the  payment  is  less  than  the  interest,  find  the 
amount  of  the  same  principal  to  the  next  payment,  or  to 
the  period  when  the  sum  of  the  payments  equals  or  exceeds 
the  interest  then  due,  and  subtract  the  sum  of  the  payments 
from  this  amount. 

Proceed  in  this  manner  with  the  balance  to  the  time 
of  settlement. 

NOTES. — i.  Tliis  method  was  early  inaugurated  by  the  Supreme 
Court  of  the  United  States,  and  is  adopted  by  New  York,  Massa. 
chusetts,  and  most  of  the  States  of  the  Union. — CJiancellor  Kent. 

2.  The  following  examples  show  the  common  forms  of  promissory 
nr  tes.    The  first  is  negotiable  by  indorsement;  the  second  by  transfer; 
the  third  is  a.  joint  note,  but  not  negotiable. 


$750. 

•  —  WASHINGTON,  D.  C.,  Jan.  "jth,  1870. 

i.  Four  months  after  date,  I  promise  to  pay  to  the  ordei 
of  George  Green,  seven  hundred  and  fifty  dollars,  with 
interest  at  6^,  for  value  received. 

HENEY 


386.  What  arc  partial  payments  ?    387.  What  is  the  United  States  rule  ? 


268  INTEREST. 

On  this  note  the  following  payments  were  indorsed. 

June  loth,  1870,  $43.     Feb.  1 7th,  1871,  $15.45.  Nov.  23d, 
1871,  $78.60.    "What  was  due  Aug.  25th,  1872  ? 

OPERATION. 

Principal,  dated  Jan.  7th,  1870,  $750.00 

Int.  to  ist  payt.  June  loth,  1870  (5  m.  3  d.)    (Art.  365),  I9-I3 

Amount,  =  769.13 

ist  payment  June  loth,  1870,  43-OO 

Remainder  or  new  principal,  =726.13 

Int.  from  ist  payt.  to  Feb.  I7th,  1871  (8  m.  7  d.),  29.89 
2!  payt.  less  than  int.  due,                                    $15.45 

Int.  on  same  prin.  to  Nov.  23d,  1871  (9  m.  6  d.),  33-4° 

Amount,  =  789.42 

3d  payt.  to  be  added  to  2d,                                    678.60  =    94-°5 

Remainder  or  new  principal,  =  695.37 

Int.  to  Aug.  25th,  1872  (9  m.  2  d.),  3I-52 

Balance  due  Aug.  25th,  1872,  =  $726.89 


$1500. 

NEW  ORLEANS,  July  ist,  1869. 

2.  Two  years  after  date,  we  promise  to  pay  to  James 
Underbill  or  bearer,  fifteen  hundred  dollars,  with  interest 
at  7$,  value  received. 

G.  H.  DENNIS  &  Co. 

Indorsements: — Received,  Jan.  5th,  1870,  $68.50.  Aug. 
8th,  1870,  $20.10.  Feb.  nth,  1871,  $100.  How  much 
was  due  at  its  maturity  ? 


$930. 

ST.  Louis,  March  $t7i,  1860. 

3.  On  demand,  we  jointly  and  severally  promise  to  pay 
J.  C.  Williams,  nine  hundred  and  thirty  dollars,  with 
interest  at  8$,  value  received.  THOMAS  BENTON. 

HENKY  VALENTINE. 

Indorsements: — Received  Oct.  loth,  1860,  $20.  Nov. 
t6th,  1861,  $250.13.  June  2oth,  1862,  $310:  what  was 
iuc  Jan.  3oth,  1863? 


PARTIAL     PAYMEXT3.  269 


MERCANTILE    METHOD. 

Sf  f .  When  notes  and  interest  accounts  payable  within  a 
year,  receive  partial  payments,  business  men  commonly 
employ  the  following  method : 

Find  the  amount  of  the  whole  debt  to  the  time  of  settle- 
ment; also  find  the  amount  of  each  payment  from  the  time 
it  was  made  to  the  time  of  settlement. 

Subtract  the  amount  of  the  payments  from  the  amount  of 
.the  debt;  the  remainder  ivill  be  the  balance  due. 

4.  A  debt  of  $720.75  was  duo  March  isth,  1870,  on 
which  the  following  payments  were  made :  April  3d,  $170 ; 
Mtiy  2oth,  $245.30;  June  17^,887.50.    How  much  was 
due  at  6%,  Sept.  5th,  1870  ? 

Principal  dated  March  isth,  1870,  $720.75 

Int.  to  settlement  (174  d.)=$72o.75  x  .029,    (Art.  367.)  *  20.90 

Amount,  Sept.  5,  '70,  =  74J'65 

ist  payt.,  $170.  Time,  155  d.  Amt ,  =$174.39 
2d  payt.,  $245.30.  Time,  108  d.  Amt.,  =  249.72 
3d  payt.,  $87.50.  Time,  80  d.  Amt.,  =  88.67 

Amt.  of  the  Payts.,  =  512.78 

Balance  due  Sept.  5th,  1870,  $228.87 

5.  Sold  goods  amounting  to  $650,  to  be  paid  Jan.  ist, 
1868.    June  loth,  received  $125;  Sept.  1 3th,  $75.50;  Oct. 
3d,  $210:  what  was  due  Dec.  3ist,  1868,  at  6%  interest? 

6.  A  note  for  $820,  dated  July  5th,  1865,  payable  in 
i  year,  at  1%  interest,  bore  the  following  indorsements : 
Jan.    loth,   1866,   received  $150;   March   2oth,  received 
$73.10;    May  5th,  received  $116;    June    15^1,  received 
$141.50:  what  was  due  at  its  maturity? 

7.  An  account  of  fnoo  due  March  3d,  received  the 
following  payments:   June    ist,  $310;   Aug.  7th,  $119; 
Oct.  1 7th,  $200:  what  was  due  on  the  27th  of  the  next 
Dec.,  allowing  7^  interest  ? 


270  INTEREST. 


CONNECTICUT    METHOD. 

389.  I.  "When  the  first  payment  is  a  year  or  more  from 
the  time  the  interest  commenced. 

Find  the  amount  of  the  principal  to  that  time.  If  the 
payment  equals  or  exceeds  the  interest  due,  subtract  it  from 
the  amount  thus  found,  and  considering  the  remainder  a  neiv 
principal,  proceed  thus  till  all  the  payments  are  absorbed. 

II.  When  a  payment  is  made  before  a  year's  interest  has 
accrued. 

Find  the  amount  of  the  principal  for  i  year;  also  if  the 
payment  equals  or  exceeds  the  interest  due,  find  its  amount 
from  the  time  it  was  made  to  the  end  of  the  year,  and  sub- 
tract this  amount  from  the  amount  of  the  principal;  and 
treat  the  remainder  as  a  new  principal. 

But  if  the  payment  be  less  than  the  interest,  subtract  the 
payment  only,  from  the  amount  of  the  principal  thus  found, 
and  proceed  as  before. 

NOTE. — If  the  settlement  is  made  in  less  than  a  year,  find  the 
amount  of  the  principal  to  the  time  of  settlement ;  also  find  the 
amount  of  the  payments  made  during  this  period  to  the  same  date, 
and  subtracting  this  amount  from  that  of  the  principal,  the  remainder 
will  be  the  balance  due. — Kirby's  Reports. 


$650. 

NEW  HAVEN,  April  izth,  1860. 

8.  On  demand,  I  promise  to  pay  to  the  order  of  George 
Selden,  six  hundred  and  fifty  dollars,  with  interest,  value 
received.  THOMAS  SAWYER. 

Indorsements: — May  ist,  1861,  received  $116.20.  Feb. 
loth,  1862,  received  $61.50,  Dec.  i2th,  1862,  received 
$12.10.  June  zoth,  1863,  received  $110:  what  was  due 
Oct.  2ist,  1863? 

388.  What  is  the  mercantile  method  ?    389.  What  is  the  Connecticut  method  ? 


PARTIAL     PAYMENTS.  271 

Principal,  dated  April  i2th,  1860,  $650.00 

Int.  to  ist  payt.  May  ist,  1861  (i  y.  19  d.),  41.06 

Amount,  May  i,  '61,  =  691.06 

ist  payt.  May  ist,  1861,  116.20 

Remainder  or  New  Prin.,  May  i,  '61,          —  574-86 

Int.  to  May  i,  '62,  or  i  y.  (2d  payt.  being  short  of  i  y.),  34-49 

Amount,  May  i,  '62,  =  609.35 

Amt.  of  2d  payt.  to  May  i,  '62  (2  m.  19  d.),  62.31 

Rem.  or  New  Prin.,  May  i,  '62,  =  547-°4 

Ami.,  May  i,  '63  (i  y.),  =  5  79-86 

3d  payt.  (being  less  than  int.  due),  draws  no  int.,  12.10 

Bern,  or  New  Prin.,  May  1/63,  =  567-76 

Amt.,  Oct.  21,  '63  (5  m.  20  d.),  —  583.85 

Amt.  of  last  payt.  to  settlement  (4  m.  i  d.),  —  112.22 

Balance  due  Oct.  21,  63,  =  $47 1.63 

NOTE. — For  additional  exercises  in   the  Connecticut  rule,  the 
ptudent  is  referred  to  Art.  387. 

VERMONT    RULE. 

390.  I-  "  When  payments  are  made  on  notes,  bills,  or  similar 
obligations,  whether  payable  on  demand  or  at  MI  specified  time, '  with 
interest,'   such  payments  shall  be  applied;  First,  to  liquidate  the 
interest  that  has  accrued   at   the  time   of  such   payments ;    and, 
secondly,  to  the  extinguishment  of  the  principal." 

II.  "  The  annual  interests  that  shall  remain  unpaid  on  notes,  bills, 
cr  similar  obligations,  whether  payable  on  demand  at  a  specified 
time,  "  with  interest  annually,"  shall  be  subject  to  simple  interest 
from  the  time  they  become  due  to  the  time  of  final  settlement." 

III.  "  If  payments  have  been  made  in  any  year,  reckoning  from  the 
time  such  annual  interest  began  to  accrue,  the  amount  of  such  pay- 
ments at  the  end  of  such  year,  with  interest  thereon  from  the  time 
of  payment,  shall  be  applied  ;  First,  to  liquidate  the  simple  interest 
that  has  accrued  from  the  unpaid  annual  interests. 

"  Secondly,  To  liquidate  the  annual  interests  that  have  become  due; 
"  Thirdly,  To  the  extinguwJiment  of  the  principal." 

391.  The  Rule  of  New  Hampshire,  when  partial  payments  ar« 
made  on  notes  "  with  interest  annually,"  is  essentially  the  same  as 
the  preceding.    But  "  where  payments  are  made  expressly  on  account 
of  interest  accruing,  but  not  then  due,  they  are  applied  when  the 
interest  falls  due,  without  interest  on  such  payments." 


272 


PABTIAL     PAYMENTS. 


$1500. 


BURLINGTON,  Feb.  x 
9.  On  demand,  I  promise  to  pay  to  the  order  of  Jam! 
Sparks,  fifteen  hundred  dollars,  with  interest  annually, 
received.  AUGUSTUS  WARREN. 


Indorsements: — Aug.  ist,  i865,received$i6o;  July  i2th. 
1866,  $125  ;  June  i8th,  1867,  $50.  Required  the  amouri 
due  Feb.  ist,  1868. 

Principal,  $1500.00 

Int.  to  Feb.  i,  '66  (i  yr.  at  6%),  90.00 

Amount,  =  i59O.o<? 

ist  payment  Aug.  i,  '65,  $160.00 

Int.  on  same  to  Feb.  i,  '66  (6  mos.),  4.80                164.80 

Remainder  or  new  principal,  1425.20 

Int.  on  same  to  Feb.  i,  '67  (i  yr.),  85.5 1 

Amount, 

2d  payment  July  12,  '66, 
Int.  on  same  to  Feb.  i,  '67  (6  m.  20  d.), 

Remainder  or  new  principal, 
Int.  on  same  to  Feb.  i,  '68  (i  yr.), 

Amount, 

3d  payment  June  18,  '67, 
Int.  on  same  to  Feb.  i,  '68  (7  m.  14  d.), 

Bal.  Feb.  ist,  1868,  =  $1412.57 


5125.00 
4.16 

129.16 

$50.00 

1.87 

-  1381.55 
82.89 

=  146444 

51-87 

$2000. 


CONCORD,  Jan.  i$th,  1869. 


10.  Two  years  after  date,  I  promise  to  pay  to  the  order 
of  Lewis  Hunt,  two  thousand  dollars,  "with  interest:" 
the  payee,  Jan.  isth,  1870,  received,  by  agreement,  $200 
on  account  of  interest  then  accruing.  What  was  due  on 
this  note  Jan.  15th,  1871,  by  the  Vt.  and  1ST.  H.  rules? 

By  the  Vt.  rule,  the  bal.=$224O— $212=^2028  )    . 
"      N.  H.  rule,     "      =$2240— $200=^2040  \ 

In  the  former,  interest  is  allowed  on  the  payment  from  its  date  to 
the  settlement ;  in  the  latter,  it  is  not 


COMPOUND    INTEREST. 

392.  Interest  may  be  compounded  annually,  semi" 
annually,  quarterly,  or  for  any  other  period  at  which  the 
interest  is  made  payable. 

393.  To  compute   Compound  Interest,  the   Principal, 
the  Rate,  and  Period  of  compounding  being  given. 

1.  What  is  the  compound  interest  of  $600  for  3  years, 
at  6%'i 

Principal,  $600.00 

Int.  for  ist  year,  $600  x  .06,  36.00 

Amt.  for  i  y.,  or  2d  prin.,  =  636.00 

Int.  for  2d  year,  $636  x  .06,  38.1  6 

Amt.  for  2  yrs.,  or  3<i  prin.,  =  674.16 

Int.  for  3d  year,  $674.16  x  .06,  4°-45 

Amt,  for  3  years,  =  714.61 

Original  principal  to  be  subtracted,  600.00 

Compound  int.  for  3  years,  =  &i  14.61 
Hence,  the 

RULE.—  I.  Find  the  amount  of  the  principal  for  the  first 
period.  Treat  this  amount  as  a  new  principal,  and  find 
the  amount  due  on  it  for  another  period,  and  so  on  through 
every  period  of  the  given  time. 

II.  Subtract  the  given  principal  from  the  last  amount, 
and  the  remainder  will  be  the  compound  interest. 

NOTES.  —  i.  If  there  are  months  or  days  after  the  last  regular 
period  at  which  the  interest  is  compounded,  find  the  interest  on  the 
amount  last  obtained  for  them,  and  add  it  to  the  same,  before  sub- 
tracting the  principal. 

2.  Compound  interest  cannot  be  collected  bylaw;  but  a  creditor 
may  receive  it,  without  incurring  the  penalty  of  usury.  Saving* 
Banks  pay  it  to  all  depositors  who  do  not  draw  their  interest 
when  due. 

2.  What  is  the  compound  ink  of  $500  for  3  yrs.  at  7%  ? 

3.  What  is  the  compound  int.  of  8750  for  4  yrs.  at 


393.  How  find  compound  interest,  when  the  principal,  rate,  and  time  are  given? 


274 


COMPOUND     INTEREST. 


4.  What  is  the  com.  int.  of  $1000  for  2  y.  7  m.  9  d.  at  6£? 

5.  What  is  the  interest  of  $1360  for  2   years  at  7^, 
compounded  semi-annually  ? 

6.  What  is  the  amount  of  $2000  for  2  years  at  ^'/com- 
pounded quarterly  ? 

COMPOUND     INTEREST    TABLE. 

Showing  the  amount  of  $i,  at  3,  4,  5,  6,  and  1%  com- 
pound  interest,  for  any  number  of  years  from  i  to  25. 


Yrs. 

9*. 

4%. 

&. 

6#. 

H6. 

I. 

1.030  ooo 

1.040  ooo 

1.050000 

1.060000 

1.07  ooo 

2. 

1.  060  900 

1.081  600 

1.  102  500 

1.123  600 

1.14490 

3- 

1.092  727 

1.124  864 

I.I57625 

1.191  016 

1.22  504 

4 

1.125  5°9 

1.169859 

I.2I5506 

1.262477 

1.31079 

5- 

1.159274 

1.216653 

1.276  282 

1.338226 

1.40  255 

6. 

1.194052 

1.265319 

1.340  096 

1.418519 

1.50073 

7- 

1.229  874 

I-3I5932 

1.407  100 

1.503  630 

1.60578 

8. 

1.266  770 

1.368569 

1-477455 

1-593  848 

I.7l8l8 

9- 

I-304773 

1.423312 

i-55I328 

1.689479 

1.83  845 

10. 

1-343  9l6 

1.480  244 

1.628  895 

1.790848 

1.96715 

ii. 

1.384234 

I-53945I 

1.710339 

1.898  299 

2.10485 

12. 

1.425  761 

i  .60  1  032 

1-795  856 

2.012  196 

2.25  219 

13- 

1.468534 

1.665  °74 

1.885  649 

2.132  928 

2.4d  984 

14- 

1.512590 

1.731  676 

1.979932 

2.26o  904 

2-57  853 

15- 

1-557  967 

1.800  944 

2.078928 

2.396558 

2-75  903 

1  6. 

1.604  7°6 

1.872981 

2.182875 

2.540352 

2.95  216 

17- 

1.652848 

1.947  900 

2.292  018 

2.692  773 

3.15881 

18. 

1-702433 

2.025  817 

2.406  619 

2-854339 

3-37  293 

19. 

1-753  5°6 

2.106  849 

2.526950 

3.025  600 

3.61  652 

20. 

1.806  in 

2.191  123 

2.653  298 

3-207  135 

3.86  968 

21. 

1.860295 

2.278768 

2-785  963 

3-399564 

4.14056 

22. 

1.916  103 

2.369919 

2.925  261 

3-603537 

4.43  040- 

23- 

i-973587 

2.464  716 

3-071524 

3-8i975o 

4-74052 

24. 

2.032  794 

2-563  304 

3.225  100 

4.048  935 

5.07  236 

25- 

2.093  778 

2.665  836 

3-386355 

4.291  871 

5-42  743 

COMPOUND     INTEHEST.  275 

394.  To  find  Compound  Interest  by  the  Table. 

•j.  Yvrhat  is  the  amount  of  $900  for  6  yrs.  at  7$,  the  int. 
Compounded  annually  ?  What  is  the  compound  interest  ? 

SOLUTION— Tabular  amt.  of  $i  for  6  yrs.  at  1%,  $1.50073 

The  principal,  90; 

Ami.  for  6  yrs.,  =  81350.65700 
The  principal  to  be  subtracted  from  amt.,  900 

Compound  Int.  for  6  yrs.,  =    $450.657 
Hence,  the 

RULE. — I.  Multiply  the  tabular  amount  of  $i  for  the  given 
time  and  rate  by  the  principal ;  the  product  will  be  the 
amount. 

II.  From  the  amount  subtract  the  principal,  and  the 
remainder  will  be  the  compound  interest. 

NOTE. — If  the  given  number  of  years  exceed  that  in  the  Table, 
find  the  amount  for  any  convenient  period,  as  half  the  given  years  ; 
then  on  this  amount  for  the  remaining  period. 

8.  What  is  the  interest  of  $800  for  9  years  at  6£,  com- 
pounded annually  ? 

9.  What  is  the  int.  of  $1100  for  12  years  at  7^,  com- 
pounded annually  ? 

10.  What  is  the  int.  of  $1305  for  16  years  at  $%,  com- 
pounded annually? 

11.  What  is  the  amount  of  $4500  for  15  years  at  4^', 
compounded  annually  ? 

12.  What  is  the  amount  of  $6000  for  25  years  at  7$, 
compounded  annually? 

13.  What  is  the  amount  of  $3800  for  30  years  at  6$, 
compound  interest? 

14.  What  is  the  compound  interest  of  $4240  at  $%  fcr 
40  years  ? 

15.  What  is  the  amount  of  $1280  for  50  years  at  -]% 
compound  interest  ? 

1 6.  What  will  $100  amount  to  in  60  years  at  6%  com- 
pound interest? 


. 


276  DISCOUNT. 


DISCOUNT. 

395.  Commercial  Discount  is  a  deduction  of  a 
certain  per  cent  from  the  price-list  of  goods,  the  face  of 
bills,  &c.,  without  regard  to  time. 

396.  True  Discount  is  the  difference  between  a 
debt  bearing  no  interest  and  its  present  ivorth. 

397.  The  Present  Worth  of  a  debt  payable  at  a 
future  time  without  interest,  is  the  sum,  which,  put  at 
legal  interest  for  the  given  time,  Avill  amount  to  the  debt. 

397,  a.  To  find  tha   Net  Proceeds  of  Commercial   Discount. 

1.  Sold  goods  marked  81560,  art  20%  discount,  on  4m., 
then  deducted  5%  for  cash.    Required  the  net  proceeds? 

ANALYSIS.  —  $1560  x  .20  =  $312.00,  and    $1560  —  $312  =  $1248. 
§1248  x  .05  =$62.40,  and  §1248—  $62.4o=$n85.6o,  net.     Hence,  the 

RULE.  —  Deduct  the  commercial  discount  from  the  list 
price,  and  from  the  remainder  take  the  cash  discount. 

2.  What  is  the  net  value  of  a  bill  of  $3500,  at  15$  dis- 
count, and  $%  additional  for  cash?  Ans.  $2826.25. 

3.  What  is  the  net  value  of  a  bill  amounting  to  85280, 
at  i  z\%  discount  ?  Ans.  $4620. 


398.  To  find  the  Present  Worth  of  a  debt,  the  Rate  and 
Time  being  given. 

1.  What  is  the  present  worth  of  $250.51,  payable  iu 
8  months  without  interest,  money  being  worth  6%  ? 

ANALYSIS.  —  The  amount  of   $i   for  8         $i.o4  =  amt.  £2 

mos.,  at  6fc,  is  $1.04  ;  therefore  $i  is  the  $1.04)8250.51  debt 
present  worth  of  $1.04,  due  in  8  mos.,  &2AO  871;    Ans 

and$250.5i-H-$i.o4=$24O.875.  Hence,  the 

RULE.  —  Divide  the  debt  by  the  amount  of  Si  for  the  gicen 
time  and  rate  ;  the  quotient  will  be  the  present  worth. 

2.  Find  the  present  worth  of  $300,  due  in   10  m.,  when 
interest  is  1%. 

395.  What  is  discount?     396.  Commercial?     True?     397.   Present  worth? 
397,  a.  How  find  net  proceeds  of  commercial  dUcotmt  ?    398.  Present  worth  ? 


DISCOUNT.  277 

3.  Find  the  present  worth  of  $500,  due  in  i  year,  when 
interest  is  8%. 

4.  What  is  the  present  worth  of  a  note  for  $1250,  pay- 
able  in  6  mos.,  interest  being  6%  ? 

5.  A  man  sold  his  farm  for  $2500  on  i  year  without 
interest:  what  is  the  present  worth  of  the  debt,  money 
being  7%? 

6.  What  is  the  present  worth  of  a  legacy  of  85000, 
payable  in  2  years,  when  interest  is  6%  ? 

399.    To   find    the    True   Discount,   the    Rate    and   Time 
being  given. 

7.  What  is  the  true  discount  at  6^  on  a  note  of  8474.03 
due  in  6  months  and  3  days  ? 

AKALYSis._The   amount    of   $i    for  $I.o3o5)§474.o3 

6  mos.  and  3  days  is  $1.0305.    (Art.  365.)  -  <?/ 

Therefore,  the  present  worth  of  the  note  8400 

is  $474.03-^1.0305,  or  $460.    (Art.  398.)  8474-03— 466  =  $I4.O3 
But,  by  definition,  the  debt,  minus  the 

present  worth,  is  the  true  discount ;   and  $474.03— $460  (present 
worth)=$i4.o3,  the  Ans.    Hence,  the 

RULE. — First  find  the  present  worth;  then  subtract  it 
from  the  debt. 

8.  (Find  the  discount  on  $2560,  due  in  7  months,  at  6fc. 

9.  Find  the  discount  on  $2819,  due  in  9  months,  at  $%. 

10.  Bought  $2375  worth  of  goods  on  6  months:  what  is 
the  present  worth  of  the  bill,  at  8%  ?  The  discount  ? 

11.  At  6%  what  is  the  present  worth  of  a  debt  of  $3860, 
half  of  which  is  due  in  3  months  and  half  in  6  months? 

12.  What  is  the  difference  between  the  int.  of  $6000  for 
i  y.  at  6%,  and  the  discount  for  i  y.  at  6^? 

13.  Bought  a  house  for  85560   on    i|  year  vithout 
interest:  what  would  be  the  discount  at  7%,  if  paid  down  ? 

14.  If  money  is  worth  7^,  which  is  preferable,  $15000 
cash,  or  $16000  payable  in  a  year  without  interest? 

399.  How  find  the  true  discount,  when  the  rate  and  time  are  given  ? 


278  BANKS    AND    BANK    DISCOUNT. 

BANKS    AND     BANK    DISCOUNT. 

400.  Banks  are  incorporated  institutions  which  deal 
in  money.    They  are  of  four  kinds:  banks  of  Deposit, 
Discount,  Circulation,  and  Savings. 

401.  A   BanJs  of  Deposit  is  one  that  receives 
money  for  safe   keeping,   subject  to   the   order  of   the 
depositor. 

A  Bank  of  Discount  is  one  that  loans  money, 
discounts  notes,  drafts,  etc. 

A  Bank  of  Circulation  is  one  that  makes  and 
issues  bills,  which  it  promises  to  pay,  on  demand. 

A  Savings  Bank  is  one  that  receives  small  sums 
on  deposit,  and  puts  them  at  interest,  for  the  benefit 
of  depositors. 

402.  Bank  Discount  is  simple  interest  paid  in 
advance. 

NOTES. — i.  A  note  or  draft  is  said  to  bo  discounted  when  the 
interest  for  the  given  time  and  rate  is  deducted  from  the  face  of  it, 
and  the  balance  paid  to  the  holder. 

2.  The  part  paid  to  the  holder  is  called  the  proceeds  or  avails  of 
the  note;  the  part  deducted,  the  dincount. 

3.  The  time  from  the  date  when  a  note  is  discounted  to  its  ma- 
turity, is  often  called  the  Term  of  Discount. 

403.  To  find  the  Bank  Discount,  the  Face  of  a  Note,  the 
Time,  and  the  Rate  being  given. 

i.  What  is  the  bank  discount  on  $450  for  4  m.  at  6%? 

ANALYSIS. — The  int.  of  $i  for  4  in.  3  d.  is  $.0205  ;  and  $450  x  .0205 
=$9.225,  the  bank  discount  required.  (Art.  380.)  Hence,  the 

EULE. — Compute  the  interest  on  the  face  of  the  note  at 
the  given  rate,  for  3  days  more  than  the  given  time. 

To  find  the  proceeds : — Subtract  the  discount  from  the 
face  of  the  note. 

400.  What  is  a  bank?  401.  A  bank  of  deposit?  Discount?  Circulation? 
Savings  bank?  402.  What  is  bank  discount?  403.  How  find  the  bank  discount, 
when  the  face  of  a  note,  the  rate  and  time  are  given  ?  How  find  the  proceeds  ? 


BANKS     AND     BANK     DISCOUNT.  579 

NOTE. — If  a  note  is  on  interest,  find  its  amount  at  maturity,  and 
taking  this  as  the  face  of  the  note,  cast  the  interest  on  it  as  above. 

)(  2.  Find  the  term  of  discount  and  proceeds  of  a  note  for 
$500,  on  90  d.,  dated  June  eth,  1873,  and  discounted  July 
3d,  1873,  at  1%.  Ans.  65  d.;  Pro.,  1493.68. 

3.  Find  the  proceeds  of  a  note  of  $730  due  in  3m.  at  6%. 

4.  Find  the  proceeds  of  a  draft  for  $1000  on  60  d.  at  6%. 

5.  Find  the  maturity,  term  of  discount,  and  proceeds  of 
a  note  of  $1740,  on  6m.,  dated  May  i,  '73,  and  dis.  Aug.  2ist, 
'? 3> at  5 %-     Ans.  Nov.  4th,  '7  3 ;  Time,  75  d. ;  Proceeds,  — . 

6.  "What  is  the  difference  between  the  true  and  bank 
discount  on  $5000  due  in  i  year  at  6%?         /  V  *J  u 

7.  A  jobber  buying  $7500  worth  of  goods  for  cash,  sold 
them  on  4  mos.  at  12^  advance,  and  got  the  note  dis- 
counted at  7%  to  pay  the  bill:  how  much  did  he  make ? 

404.  To  find  the  Face  of  a  Note,  that  the  proceeds  at  Bank 
Discount  shall  be  a  specified  sum,  the  R.  and  T.  being  given. 

8.  For  what  sum  must  a  note  be  drawn  on  6  mos.  that 
at  (>%  bank  discount,  the  proceeds  shall  be  $500  ? 

ANALYSIS. — The  bank  discount  of  $i  for  6  mos.  3  d.  at  6^  is 
$.0305 ;  consequently  the  proceeds  are  $.9695.  Now  as  $.9695  are 
the  proceeds  of  551,  $500  must  be  the  proceeds  of  as  many  dollars  as 
$.9695  is  contained  times  in  $500;  and  $5oo-5-$.9695=$5i5.729,  the 
face  of  the  note.  Hence,  tho 

RULE. — Divide  the  given  proceeds  ly  the  proceeds  of  $i 
for  the  given  time  and  r^te. 

9.  What  must  be  the  face  of  a  note  on  4  mos.  that  when 
discounted  Bt  1%  the  procoeds  may  be  $750  ? 

10.  What  was  the  face  of  a  note  on  60  days,  the  pro- 
ceeds of  which  being  discounted  at  5$,  were  81565  ? 

11.  If  the  avails  are  $2165.45,  the  time  4  mos.,  and  the 
rate  of  bank  discount  8$,  what  must  be  the  face  of  the  note? 

12.  Bought  a  house  for  $7350  cash :  how  large  a  note  on 
4  mos.  must  I  have  discounted  at  bank  6%  to  pay  this  sum? 


404.  How  find  how  large  to  make  a  note  to  raise  a  specified  sam,  when  the  rato 
and  time  are  given  » 


280  STOCK    INVESTMENTS. 

STOCK   INVESTMENTS. 

405.  Stocks  are  the  funds  or  capital  of  incorporated 
companies. 

An  Incorporated  Company  is  an  association 
authorized  by  law,  to  transact  business. 

NOTE. — Stocks  are  divided  into  equal  parts  called  shares,  and  tlio 
owners  of  the  shares  are  called  stockholders.  These  shares  vary  from 
$25  to  $500  or  $1000.  They  are  commonly  $100  each,  and  will  be 
so  considered  in  the  following  exercises,  unless  otherwise  stated. 

406.  Certificates  of  Stock  are  written  statements, 
specifying  the  number  of  shares  to  which  holders  are 
entitled.     They  are  often  called  scrip. 

407.  The  JPar  value  of  stock  is  the  sum  named  on 
the  face  of  the  scrip,  and  is  thence  called  its  nominal  value. 

The  Market  value  is  the  sum  for  which  it  sells. 

NOTES. — i.  When  shares  sell  for  their  nominal  value,  they  are  at 
par;  when  they  sell  for  more,  they  are  above  par,  or  at  &  premium; 
when  they  sell  for  less,  they  are  below  par,  or  at  a  discount. 

2.  When  stocks  sell  at  par,  they  are  often  quoted  at  100 ;  when  8  % 
above  par,  at  108  ;  when  %%  beloio  par,  at  92.  The  term  pur,  Latin, 
signifies  equal ;  hence,  to  be  at  par,  is  to  be  on  an  equality. 

408.  The  Gross  Earnings  of  a  Company  are  its  entire 

receipts. 

The  Net  Earnings  are  the  sums  left  after  deducting  all 
expenses. 

409.  Instalments  are  portions  of  the  capital  paid  by  the 
stockholders  at  different  times. 

410.  Dividends  are  portions  of  the  earnings  distributed  among 
the  stockholders.     They  are  usually  made  at  stated  periods ;  as, 
annually ;  semi-annually,  etc. 

411.  A  Bond  is  a  writing  under  seal,  by  which  a 
party  binds  himself  to  pay  the  holder  a  certain  sum,  at  or 
before  a  specified  time. 

405.  What  are  stocks  ?  An  incorporated  company  ?  Note.  Into  what  are 
Blocks  divided  ?  What  are  stockholders  ?  406.  What  are  certificates  of  stock  ? 

407.  What  is  the  par  value  of  stock?    The  market  value?     Note.  When  are 
stocks  at  par?     Above  par?    Below  par?    The  meaning  of  the  term  par? 

408.  What  are  the  gross  earnings  of  a  company  ?    The  net  earnings  ?    405.  Whul 
ore  instalments ?    410.  Dividends?    411.    What  is  a  bond? 


BTOCK    INVESTMENTS.  283 

UNITED    STATES   BONDS. 

412.  United  States  Bonds  are  those  issued  by 
Government,  and  are  divided  into  two  classes :  those  pay- 
able at  a  given  date,  and  those  payable  within  the  limits 
of  two  given  dates,  at  the  option  of  the  Government. 

NOTE. — The  former  are  designated  by  a  combination  of  the 
numerals,  which  express  the  rate  of  interest  they  bear,  and  the  year 
they  become  due  ;  as  "  6s  of  '81." 

The  latter  are  designated  by  combining  the  numerals  expressing 
the  two  dates  between  which  they  are  to  be  paid  ;  as  "  5-203." 

413.  A  Coupon  is  a  certificate  of  interest  attached  to 
a  bond,  which,  on  the  payment  of  the  interest,  is  cut  off 
and  delivered  to  the  payor. 

414.  The  principal  U.  S.  bonds  are  the  following  : 

1.  "6s  of '8 1,"  bearing  6$  interest,  and  payable  in  1881. 

2.  "  5-203,"  bearing  6%  interest,  and  payable  in  not  less  than  5  or 
more  than  20  years  from  their  date,  at  the  pleasure  of  the  Govern- 
ment.    Interest  paid  semi-annually  in  gold. 

3.  "  10-403,'    bearing  5  %  interest,  redeemable  after  10  years  from 
their  date,  interest  semi-annually  in  gold. 

4.  "  53  of  '81,"  bearing  5  %  interest,  redeemable  after  1881,  interest 
paid  quarterly  in  gold. 

5.  "4>s  of  "86,"  bearing  $\%  interest,  redeemable  after  1886,  the 
interest  paid  quarterly  in  gold. 

6.  "43  of  IQOI,"  bearing  4%  interest,  the  principal  payable  after 
1901,  the  interest  paid  quarterly  in  gold. 

415.  State,  City,  Railroad  Bonds,  etc.,  are  payable  at  a  specified 
time,  and  are  designated  by  annexing  the  numeral  denoting  the  rate 
of  interest  they  bear,  to  the  name  of  the  State,  etc.,  by  which  they 
are  issued  ;  as,  New  York  6s ;  Georgia  73. 

416.  Computations  in  stocks  and  bonds  are  founded 
upon  the  principles  of  percentage;  the  par  value  being 
the  base,  the  per  cent  premium  the  rate,  the  premium,  etc., 
the  percentage,  and  the  market  value  the  amount. 

412.  What  are  U.  S.  bonds?  How  many  classes?  Note.  How  are  the  former 
d?sia^ate:l  ?  The  latter  ?  413.  What  is  a  coupon  ?  414.  What  is  meant  by  U.  S, 
teof'si?  By  U.  S.  s-208  ?  By  U.  S.  10-408?  By  new  U.  S.  5*  of '81  ? 


282  STOCK    INVESTMENTS 

NOTE.— The  comparative  profit  of  investments  in  U.  S.  Bonds  de- 
pends upon  their  market  value,  the  rate  of  interest  they  bear,  and 
the  premium  on  gold.  That  of  Railroad  and  other  Stocks-«pon  their 
market  value,  and  the  per  cent  of  their  dividends 

•v  4- 

417.  To  find  the  Premium,  Discount,  Instalment,  or 
Dividend,  the  Par  Value  and  the  Rate  being  given. 

1.  "What  is  the  premium  on  20  shares  of  the  New  Orleans 
National  Bank,  at  8$  ? 

ANALYSIS. — 208. =$2000;  and  $2000  x  .08 =$160,  Ana.  Hence,  the 

RULE. — Multiply  the  par  value  by  the  rate,  expressed 
decimally.  (Art.  336.) 

2.  What  is  the  discount  on  27  shares  of  Michigan 
Central  Eailroad  at  g%  ? 

3.  What  is  the  premium  on  three  $1000  IT.  S.  6s  of  '81, 
•when  they  stand  at  \i\%  above  par  ? 

4.  The  Maryland  Coal  Company  called  for  an  instalment 
of  15%:  what  did  a  man  pay  who  owned  35  shares? 

5.  What  is  the  discount  on  $4000  Tennessee  6s,  at  IO;T/? 

6.  The  Virginia  Manufacturing   Company  declared  a 
dividend  of  17^:  to  what  sum  were  28  shares  entitled? 

418.  To  find  the  Market  Value  of  Stocks  and  Bonds,  the 

Rate  and  Nominal  Value  being  given. 

7.  What  is  the  market  value  of  45  shares  of  New  York 
Central,  at  A,%  premium  ? 

ANALYSIS. — 459. =§4500;  and  $4500x1.04  (i  +  the  rate)=$468o, 
the  value  required.  Hence,  the 

RULE. — Multiply  the  nominal  value  by  i  plus  or  minus 
the  rate. 

Or,  multiply  the  market  value  of  i  share  by  the  number 
of  shares. 

NOTE. — In  finding  the  net  value  of  stocks,  the  brokerage,  postage 
stamps,  and  other  expenses  must  be  deducted  from  the  market  value. 

8.  What  is  the  worth  of  814000  of  Kentucky  6s,  at  92  ? 

417.  How  find  the  premium,  etc.,  when  the  par  value  and  rate  are  given? 


STOCK    INVESTMENTS.  283 

9.  What  is  the  worth  of   an  investment  of  $9500  in 
tJ.  S.  5-203,  at  \z>\%  premium  ? 

10.  What  is  the  net  value  of  58  shares  of  New  Jersey 
Central,  at  no;  deducting  the  express  and  other  charges 
$1.89,  and  brokerage  at  \%  ? 

=n»i.  What  will  be  realized  from  $7500  Texas  75,  at  15$ 
discount,  deducting  the  brokerage  at  %f0,  and  $11.39  f°r 
postage  and  other  charges. 

12.  What  will  $10000  U.  S.  5-203  cost  at  ^\%  premium, 
adding  brokerage  at  \%,  and  other  expenses  $^.37^-? 

13.  What  is  the  value  in  currency  of  $15750  gold  coin, 
the  premium  being  22-^? 

14.  A  person  has  $10000  U.  S.  5-203:  what  will  he  re- 


ceive annually  in  currency,  when  gold  is  1 2%  premium  ? 

419.    To  find  the  Rate,  the   Par  Value,  and  the  Dividend, 
Premium,  or  Discount  being  given. 

15.  The  capital  stock  of  a  company  is  fiooooo;   its 
gross  earnings  for  the  year  are  $34500,  and  its  expenses 
$13500 :  deducting  from  its  net  earnings  $1000  as-  surplus, 
what  per  cent  dividend  can  the  company  make  ? 

ANALYSIS. — $13500  +  $1000  =  $14500,  and  $34500  —  $14500  = 
$20000,  the  net  earnings.  Now  20000 -H$IOOOOO=. 20  or  20$,  the 
rate  required.  Hence,  the 

RULE. — Divide  the  premium  or  discount,  as  the  case  may 
be,  by  the  par  value.  (Art.  339.) 

16.  Paid  $750  premium  for  50  shares  of  bank  stock: 
what  was  the  per  cent? 

17.  The  discount  on  100  shares  of  the  Pacific  Eailroad 
is  $625  :  what  is  the  per  cent  below  par? 

1 8.  The  net  earnings  of  a  company  with  a  capital  of 

$480000  are  $35000 ;   reserving  $3000  as   surplus,  what 
per  cent  dividend  can  they  declare  ? 

4:8.  How  find  the  market  value  when  the  rate  and  nominal  valne  are  given? 
419.  How  fiud  the  rate,  when  the  premium,  discount,  or  dividend  are  given? 


284  STOCK    INVESTMENTS. 

420.  To  find  how  much  Stock  can  be  bought  for  a  specified 
sum,  the  Rate  of  Premium  or  Discount  being  given. 

19.  How  much  of  Kansas  6s,  at  20$  discount,  can  be 
bought  for  $5200  ? 

ANALYSIS. — $5200-=-. 80  (val.  of  $i  8tock)=$6soo,  Ans.    Hence,  the 

RULE. — Divide  the  sum  to  be  invested  by  i  plus  or  minus 
the  rate,  as  the  case  may  be.  (Art.  349.) 

Or,  divide  the  given  sum  by  the  market  value  q/*$i  of  stock. 

20.  How  many  $100  IT.  S.  10-403,  at  $%  premium,  can 
be  bought  for  $4200  ? 

2 1.  What  amount  of  Virginia  6s,  at  90,  can  be  purchased 
for  $10800? 

421.  To  find  what  Sum  must  be  invested  in  Bonds  to  realize 
a  given  income,  the  Cost  and  Rate  of  Interest  being  given. 

22.  What  sum  must  be  invested  in  Missouri  6s  at  90,  to 
realize  an  income  of  $1800  annually? 

ANALYSIS. — At  6%  the  income  of  $i  is  6  cts. ;  and  $1800-5- .06= 
$30000,  the  nominal  value  of  the  bonds.  Again,  at  .90,  $30000  of  bonds 
will  cost  .90  times  f  30000=$ 27000,  the  sum  required.  Hence,  the 

RULE. — I.  Divide  the  given  income  by  the  annual  interest 
9/  $i  of  bonds ;  the  quotient  will  be  the  nominal  value  of 
the  bonds. 

II.  Multiply  the  nominal  value  by  the  market  value  of$i 
of  bonds;  the  product  will  be  the  sum  required. 

23.  What  sum  must  be  invested  in  IT.  S.  5-205,  at  106, 
to  yield  an  annual  income  of  $2500  in  gold? 

24.  How  much  must  one  invest  in  Wisconsin  8s,  at  95, 
to  receive  an  annual  income  of  $3000  ?  Ans.  $35625. 

25.  How  much  must  be  invested  in  Mississippi  6s,  at  80, 
to  yield  an  annual  interest  of  $4200  ? 

420.  How  find  the  quantity  of  stock  that  can  be  bought  for  a  specified  sum, 
when  the  rate  is  given  ? 


EXCHANGE. 

422.  Exchange  is  a  method  of  making  payments 
between  distant  places  by  Bills  of  Exchange. 

423.  A  Bill  of  Exchange  is  an  order  or  draft 
directing  one  person  to  pay  another  a  certain  sum  at  a 
specified  time. 

2vOTE. — The  person  who  signs  the  bill  is  called  the  drawer  or 
maker ;  the  one  to  whom  it  is  addressed,  the  drawee;  the  one  to 
whom  it  is  to  be  paid,  the  payee  ;  t'lie  one  who  sends  it,  the  remitter. 

424.  Bills  of  Exchange  are  Domestic  or  Foreign* 
Domestic  Hills  are  those  payable  in  the  country 

where  they  are  drawn,  and  are  commonly  called  Drafts. 

Foreign  Bills  are  those  drawn  in  one  country  and 
payable  in  another. 

425.  A  Sight  Bill  is  one  payable  on  its  presentation. 

A  Time  Bill  is  one  payable  at  a  specified  time  after 
its  date,  or  presentation. 

426.  The  Par  of  Exchange  is  the  standard  by 
which  the  value  of  the  currency  of  different  countries  is 
compared,  and  is  either  intrinsic  or  nominal. 

An  Intrinsic  Par  is  a  standard  having  a  real  and 
fixed  value  represented  by  gold  or  silver  coin. 

A  Nominal  far  is  a  conventional  standard,  having 
any  assumed  value  which  convenience  may  suggest. 

427.  When  the  market  price  of  bills  is  the  same  as  the 
face,  they  are  at  par;  when  it  exceeds  the  face,  they  are 
above  par,  or  at  a  premium  ;  when  it  is  less  than  the  face, 
they  are  below  par,  or  at  a  discount     (Art.  407,  n.} 

NOTES. — i.  The  fluctuation  in  the  price  of  bills  from  their  par 
value,  is  called  the  Course  of  Exchange. 

422.  What  is  exchange?  423.  A  bill  of  exchange?  424.  What  arc  domestic 
bills?  Foreign  hills T  425.  What  are  eight  hills?  Time  bills?  426.  What  i» 
the  par  of  exchange  ?  An  intrinsic  par  ?  A  nominal  par  ? 


286  DOMESTIC     EXCHANGE. 

2.  The  rate  of  Exchange  between  two  places  or  countries  depends 
upon  the  circumstances  of  trade.  If  the  trade  between  New  York 
and  New  Orleans  is  equal,  exchange  is  at  par.  If  the  former  owes 
the  latter,  the  demand  for  drafts  on  New  Orleans  is  greater  than  the 
supply;  hence  they  are  above  par  in  New  York.  If  the  latter  owes 
the  former,  the  demand  for  drafts  is  less  than  the  supply,  con- 
sequently drafts  on  New  Orleans  are  below  par. 

428.  An  acceptance  of  a  bill  or  draft  is  an  engage- 
ment to  pay  it  according  to  its  conditions.     To  show  this, 
it  is  customary  for  the  drawee  to  write  the  word  accepted 
across  the  face  of  the  bill,  with  the  date  and  his  name. 

NOTE.  —  Bills  of  Exchange  are  negotiable  like  promissory  notes, 
and  the  laws  respecting  their  indorsement,  collection,  protest,  etc., 
are  essentially  the  same. 

DOMESTIC    OR    INLAND    EXCHANGE. 

429.  To  find  the  Cost  of  a   Draft,  its   Face  and  the  Rate 

of  Exchange  being  given. 

i.  "What  is  the  cost  of  a  sight  draft  on  New  York,  for 

$4500,  at  z\%  premium  ? 

ANALYSIS.  —  At  z\%  premium,  a  draft  of  $i  will  $4500 

cost  $1  +  2^  cts.=$i.o25,  or  1.025  times  the  draft.  1.025 

Hence,  the  cost  of  $4500  draft  will  be  1.025  times 


T, 

*.    ,  ..  I  2.>»O 

its  face  ;  and  $4500  x  1.025  =  $4612.50, 


2.  What  is  the  cost  of  a  sight  draft  on  St.  Louis  of 
$5740,  at  4%  discount? 

ANALYSIS.  —  At  4^  discount,  a  draft  of  $i  will  $5740 

cost   $1—4  cents=$O96,  or  .96  times  the  draft  .06 

Therefore,  the  cost  of  $5740  draft  will  be  .96  times     &_  4 

Its  face  ;  and  $5740  x  .96  =$5  5  10.40,    Hence,  the 

RULE.  —  Multiply  the  face  of  the  draft  ly  the  cost  of$i 
of  draft,  expressed  decimally. 

427.  Whan  are  bills  at  par  ?  When  above  par  ?  When  below  ?  Note.  What 
is  the  fluctuation  in  the  price  of  exchange  called  ?  Upon  what  does  the  rate  of 
exchange  depend  ?  Explain  ?  428.  The  acceptance  of  a  bill  ?  429.  How  find  tho 
<:ost  of  a  draft,  when  the  face  and  rate  are  given  ? 


DOMESTIC     EXCHANGE.  287 

NOTES.  —  i.  When  payable  at  sight,  the  worth  of  $i  of  draft  is  $i 
plus  or  minus  the  rate  of  exchange. 

2.  On  time  drafts,  both  the  exchange  and  the  bank  discount  are 
computed  on  tbeir  face.  Dealers  in  exchange,  however,  make  but, 
one  computation  ;  the  rate  for  time  drafts  being  enough  less  than 
night  drafts  to  allow  for  the  bank  discount. 

.V  merchant  in  Galveston  bought  a  draft  of  62000  on 
Philadelphia  at  60  days  sight:  what  was  the  draft  worth, 
the  premium  being  z\%->  an(l  the  bank  discount  (F%.  * 

3.  What  cost  a  sight  draft  of  $3560,  at  2%  discount? 

4.  Eequired  the  worth  of  a  draft  for  $4250  on  Chicago, 
at  90  days,  sight  drafts  being  \%  discount  and  interest  1%. 

5.  The  Bank  of  New  York  having  declared  a  dividend 
of  4%,  a  stockholder  living  in  Savannah  drew  on  the  bank 
for  his  dividend  on  50  shares,  and  sold  the  draft  at  \\% 
premium  :  how  much  did  he  realize  from  the  dividend  ? 

430.  To  find  the  Face  of  a   Draft,  the  Cost  and   Rate  of 
Exchange  being  given. 

7.  Bought  a  draft  in  Omaha  on  New  York,  payable  in 
90  days,  for  $3043.50,  exchange  being  $%  premium,  and 
interest  6C/0  :  what  was  the  face  of  the  draft  ? 

ANALYSIS.—  The  cost  of  $r  of  |x  _l_  ,^—jjj.j  o- 

draft  is  $i  plus  the  rate,  minus     j)ig  on  $j  for 
the  bank  discount   for  the  time.  ^          „  ^ 

Now   Si  +  3     =  *i.Q3  ;    the   bank  °0st  °f  $' 


discount  on  $i  for  93  d.  is  $0.0155, 

and    $1.03  -  $0.0155  =  $1.0145.        Face  of  dft.=  $3000. 

Now  if  $1.0145  w'll  buy  $x   °f 

draft,  $3043.50  will  buy  a  draft  of  as  many  dollars  as  1.0145  is  con- 
tained times  in  $3043.50.  $3O43.5o-=-i.oi45  =  $3Ooo,  the  draft  re- 
quired. Hence,  the 

RULE.  —  Divide  the  cost  of  the  draft  by  the  cost  of  $i 
of  draft,  expressed  decimally. 

8.  What  is  the  face  of  a  sight  draft  purchased  for  $1250, 
the  premium  being  2  \%  ? 

430.  How  find  the  face  of  a  draft,  the  cost  and  rate  twin?  given? 


288  FOBEIGN     EXCHANGE. 

_--9.  What  is  the  face  of  a  draft  at  60  days  sight,  purchased 
for  $1500,  when  interest  is  &%,  and  the  premium  2c/ct 

10.  What  is  the  face  of  a  sight  draft,  purchased  for 
82500,  the  discount  being  $\%  ? 

__.!  i.  What  is  the  face  of  a  draft  on  4  in.,  bought  for  $3600, 
the  int.  being  d%,  and  exchange  2%  discount? 

12.  A  merchant  of  Natchez  sold  a  draft  on  Boston  at 
r£^  prem.,  for  $3806.25  :  what  was  the  face  of  the  draft  ? 

FOREIGN    EXCHANGE. 

431.  A  Foreign  Bill  of  Exchange  is  a  Bill 
drawn  in  one  country  and  payable  in  another. 

A  Set  of  Exchange  consists  of  three  bills  of  the 
same  date  and  tenor,  distinguished  as  the  First,  Second, 
and  Third  of  exchange.  They  are  sent  by  different  mails, 
in  order  to  save  time  in  case  of  miscarriage.  When  one  is 
paid,  the  others  are  void. 

432.  The  Legal  far  of  Exchange  between 
Great  Britain  and  the  United  States,  is  $4.8665  gold  to  the 
pound  sterling.* 

433.  To  find  the  Cost  of  a   Bill   on   England,  the   Face   and 
the  Rate  of  Exchange  being  given. 

i.  What  is  the  cost  of  the  following  bill  on  London,  at 
$4.8665  to  the  £  sterling? 


i2s. 


NEW  YORK,  July  4^,  1874. 


At  sight  of  this  first  of  exchange  (the  second  and  third 
of  the  same  date  and  tenor  unpaid),  pay  to  the  order  of 
Henry  Crosby,  three  hundred  and  fifty-four  pounds,  twelve 
shillings  sterling,  value  received,  and  charge  the  same  to 
the  account  of  0.  J.  KING  &  Co. 

To  GEORGE  PEABODY,  Esq.,  London. 

431.  What  is  a  foreign  bill  of  exchange  ?  What  is  the  legal  par  of  exchange 
with  Great  Britain  T 

*  Act  of  Congress,  March  sd,  1873—  To  take  effect  Jan.  ist,  1874. 


FOREIGN     EXCHANGE.  289 

ANALYSIS.  —  Reducing  the  given  shillings  to  OPERATION. 

the  decimal  of  a  pound,  £354,  128.  =  £354.6.  4.8665 

(Art.  295.)    Now  if  £i  is  worth  $4.8665,  £354.6  _  354-6 

are  worth  354  6  times  as  much;  and  $4.8665  x  Ans.  $1725.661 
354.6=$i725.66i,  the  cost  of  the  bill. 

2.  What  is  the  value  of  a  bill  on  England  for  £436,  53.  6d., 
at  $4.85^  to  the  £  sterling? 

ANALYSIS.  —  £436,  59.  6d.=  £436.275,  and  the  OPERATION. 

market  value  of  exchange  is  $4.8525  to  the  £.  $4.8525 

Now  $4.  8525  X436.275=$2ii7.o2,  the  cost  of  the  436.275 

bill.    Hence,  the  Ans.  ' 


RULE.  —  Multiply  the  market  value  of  £i  sterling  by  the 
face  of  the  bill  ;  the  product  will  be  its  value  in  dollars 
and  cents. 

NOTES.  —  i.  If  there  are  shillings  and  pence  in  the  given  bill, 
they  should  be  reduced  to  the  decimal  of  a  pound.  (Art.  295.) 

2.  Bills  on  Great  Britain  are  drawn  in  Sterling  money. 

3.  The  New  Par  of  Sterling  Exchange  $4.8665,  is  the  intrinsic 
value  of  the  Sovereign  or  pound  sterling,  as  estimated  at  the  United 
States  Mint,  and  is  9^  %  greater  than  the  old  par. 

4.  The  Old  Par,  which  assumed  the  value  of  the  £  sterling  to  be 
$4.44^,  is  abolished  by  law,  and  all  contracts  based  upon  it  after 
January  ist,  1874,  are  null  and  void. 

3.  What  is  the  cost  of  a  bill  on  Dublin  for  £38:1,  at 
$4.87!  to  the  £  sterling  ? 

4.  What  is  the  cost  in  currency  of  a  bill  on  England  for 
£750,  exchange  being  at  par,  and  gold  33$$  premium? 

5.  B  owes  a  merchant  in  Liverpool  £1500;  exchange  is 
64.93!-;  to  transmit  coin  will  cost  2$!  insurance  and  freight; 
and  when  delivered  its  commercial  value  is  $4.80  to  a 
pound:  which  is  the  cheaper,  to  buy  a  bill,  or  send  the 
gold  ?    How  much  ? 

6.  A  New  Orleans  merchant  consigned  568  bales  of  cot- 
ton, weighing  450  Ibs.  apiece,  to  his  agent  in  Liverpool; 
the  agent  paid  id.  a  pound  freight,  and  sold  it  at  i2d.  a 

433.  How  find  the  cost  of  a  bill  on  England,  when  the  face  and  rate  are  given? 


290  FOREIGN     EXCHANGE. 

• 

pound ;  he  charged  z\%  commission,  and  £8,  6s.  for  stor- 
age: what  did  the  merchant  realize  for  his  cotton,  ex- 
change on  the  net  proceeds  being  $4.91^  to  the  £? 

434.    To  find  Ihe  Face  of  a   Bill  on  England,  the  Cost  and 
the  Rate  of  Exchange  being  given. 

7.  What  is  the  face  of  a  bill  on  England  which  cost 
$1725.72,  exchange  being  at  par,  or  $4.8665  to  the  £? 

ANALYSIS. — Since  $4.8665  buys  £i  of          *    g/c/c - ^172 c  72 
bill,  $1725.72  will  buy  as  many  pounds  as 
$4.8665  are  times  in  $1725.72,  or  £354.612 
=£354,  I2s.  2fd.    Hence,  the  -Ans.  £354,  J2S.  2|d. 

RULE. — Divide  the  cost  of  the  bill  ~by  the  market  value 
of£i  sterling  ;  the  quotient  will  be  the  face  of  the  bill 

8.  What  is  the  face  of  a  bill  on  London  which  cost 
$2500,  exchange  being  $4.88^  to  the  £  ? 

3.  What  amount  of  exchange  on  Dublin  can  be  obtained 
for  $3750,  at  $4.84^  to  the  £  sterling? 

10.  A  merchant  in  Charleston  paid  $5000  for  a  bill  on 
London,  at  $4.87^  to  the  £:  what  was  the  face  of  the  bill  ? 

11.  Paid  $7500  for  a  bill  on  Manchester:  what  was  the 
face  of  it,  exchange  being  $4.86^  to  the  £? 

434,  a.  In  quoting  Exchange  on  Foreign  Countries,  the  gen- 
eral rule  is  to  quote  the  money  of  one  country  against  the  money 
of  other  countries.  Thus,  exchange  is  quoted 

On  Austria,  in  cents  to  the  Florin  (silver)r=$o.476. 

On  Frankfort,  in  cents  to  the  Florin  (gold) =$0.4 165. 

On  the  German  Empire,  in  cents  to  the  Mark  (gold)=$o  2382. 

On  North  Germany,  in  cents  to  the  new  Thaler  (silver)=; $0.714. 

On  Russia,  in  cents  to  the  Rouble  (silver) =$0.77 17. 

NOTE. — Bills  of  Exchange  between  the  United  States  and  foreign 
countries  are  generally  drawn  on  some  of  the  great  commercial 
centers ;  as  London,  Paris,  Frankfort,  Amsterdam,  etc. 

434.  How  find  the  face  of  a  bill  on  England,  when  the  cost  and  rate  are  given  ? 


FOREIGN"     EXCHANGE.  291 

<• 

435.  Sills  on  France  are  drawn  in  French  cur- 
rency, and  are  calculated  at  so  many  francs  and  centimes 
to  a  dollar. 

NOTE.— Centimes  are  commonly  written  as  decimals  of  a  franc 
Thus  5  francs  and  23  centimes  are  written  5.23  francs.  (Art.  224,  n.\ 

436.  To  find  the    Cost  of  a   Bill   on   France,  the   Face   ana 

Rate  of  Exchange  being  given. 

12.  What  is  the  cost  of  a  bill  on  Paris  of  1500  francs, 
exchange  being  5.25^  to  a  dollar  ? 

ANALYSIS. — Since  5.25  francs  will  buy  a  5.25  fr.)  1500.00  fr. 
bill  of  $i,  1500  francs  will  buy  a  bill  of  as  <  ~&2&z  71  4- 

many  dollars  as  5.25  is  contained   times  in 
1500;  and  1500-7-5.25=1285.71,  the  cost  required.     Hence,  the 

RULE. — Divide  the  face  of  the  till  by  the  number  of  franca 
to  $i  exchange. 

13.  What  is  the  cost  of  a  bill  of  3500  francs  on  Havre, 
exchange  being  5.18  francs  to  a  dollar  ? 

437.  To  find  the  Face  of  a  Bill   on  France,  the  Cost  and 

the  Rate  of  Exchange  being  given. 

14.  A  traveler  paid  $300  for  a  bill  on  Paris;  exchange 
being  5.16  francs  to  $i :  what  was  the  face  of  the  b'ill  ? 

ANALYSIS. — If  $i  will  buy  5.16  francs,  $300  will  buy  300  times  as 
many  ;  and  5.16  fr.  x  300=1548  francs,  the  face  of  the  bill  required. 
Hence,  the 

RTJLE. — Multiply  the  number  of  francs  to  $i  exchange  b$ 
the  cost  of  the  bill. 

15.  Paid  $2500   for  a  bill  on  Lyons,  exchange  being 
5.22  francs  to  £i :  what  was  the  face  of  the  bill  ? 

1 6.  A  merchant  paid  $3150  for  a  bill  on  Paris,  exchange 
being  5.23  francs  to  $i :  what  was  the  face  of  the  bill  ? 

435.  How  is  exchange  on  France  calculated  f  436.  How  find  the  cost  of  a  bill 
on  France,  when  the  face  and  rate  are  given  f  437.  How  find  the  face,  when  the 
cost  and  rate  are  given  ? 


ISTSURA^CE. 

438.  Insurance  is  indemnity  for  loss.    It  is  dis- 
tinguished by  different  names,  according  to  the  cause  of 
the  loss,  or  the  object  insured. 

439.  Fire  Insurance,  is  indemnity  for  loss  by  fire. 
Marine  Insurance,  for  loss  by  sea. 

Life  Insurance,  for  the  loss  of  life. 
Accident  Insurance,  for  personal  casualties. 
Health  Insurance,  for  personal  sickness. 
Stock  Insurance,  for  the  loss  of  cattle,  horses,  etc. 

NOTES. — i.   The  party  who  undertakes  the  risk  is  called  the 
Insurer  or  Underwriter. 
2.  The  party  protected  by  the  insurance  is  called  the  Insured. 

440.  A  Policy  is  a  writing  containing  the  evidence 
and  terms  of  insurance. 

The  Premium,  is  the  sum  paid  for  insurance. 

NOTE. — The  business  of  insurance  is  carried  on  chiefly  by  In- 
corporated Companies.  Sometimes,  however,  it  is  undertaken  by 
individuals,  and  is  then  called  out-door-insurance. 

441.  Insurance  Companies  are  of  two  kinds :  Stock  Companies, 
and  Mutual  Companies. 

A  Stock  Insurance  Company  is  one  which  has  a  paid  up 
capital,  and  divides  the  profit  and  loss  among  its  stockholders. 

A  Mutual  Insurance  Company  is  one  in  which  the  losses 
are  shared  by  the  parties  insured. 

442.  Premiums  are  computed  at  a  certain  per  cent  of 
the  sum  insured,  and  the  operations  are  similar  to  those 
in  Percentage. 

NOTE. — Policies  are  renewed  annually,  or  at  stated  periods,  and 
the  premium  is  paid  in  advance.  In  this  respect  insurance  differs 
from  commission,  etc.,  which  have  no  reference  to  time. 

438.  What  is  insurance?  439.  Fire  insurance?  Marine?  Life?  Accident? 
Health?  Stock?  440.  What  is  a  policy  ?  The  premium ?  441.  How  many  kinds 
of  insurance  companies'?  A  stuck  insurance  company?  A  mutual?  442.  How 
ere  premiums  computed  ? 


INSUEANCE.  293 

FIRE    AND     MARINE    INSURANCE. 

443.  To  find  the  Premium,  the  Sum  insured  and  the  Rate 
for  the  period  being  given. 

1.  What  premium  must  I  pay  per  annum  for  insuring 
$1750  on  my  house  and  furniture,  at  \%'t 

ANALYSIS. — $1750  x  .005  (rate)=$8.75,  Ans.    Hence,  the 
RULE. — Multiply  the  sum  insured  by  the  rate.   (Art.  336.) 
NOTE. — The  rate  of  insurance  is  sometimes  stated  at  a  certain 
number  of  cents  on  $100.     In  such  cases  the  rate  should  be  reduced 
lo  the  decimal  of  $i  before  multiplying.    Thus,  if  the  rate  is  25  cents 
on  $100,  the  multiplier  is  written  .0025.     (Art.  295.) 

2.  At  35  cents  on  $100,  what  is  the  insurance  of  $1900? 

3.  At  i-^fc,  what  is  the  premium  on  $2560? 

4.  At  z\%,  what  is  the  premium  on  $3750? 

5.  At  25  cents  on  $100,  what  is  the  premium  on  $4280? 

6.  At  50  cents  on  $100,  what  is  the  premium  on  $5000? 

7.  At  3^',  what  is  the  cost  of  insuring  $6175  ? 

8.  What  is  the  premium  for  insuring  a  ship  and  cargo 
\  ulued  at  $35000,  at  2\r/c  ? 

9.  What  is  the  cost  of  insuring  a  factory  and  its  con- 
d'lits,  valued  at  $48250,  at  $\%,  including  $i-J  for  policy? 

444.  To  find  what  Sum  must  be  insured  to  cover  both  the 
Property  and  Premium,  the  Rate  being  given. 

10.  For  what  must  a  factory  worth  $20709  be  insured, 
to  cover  the  property  and  the  premium  of  2\%  ? 

ANALYSIS. — The  sum  to  be  insured  includes  the  property  plus  the 
premium.  But  the  property  is  160%  of  itself,  and  the  premium  is 
z}%  of  that  sum  ;  therefore  $20709  =  100 %  —21%,  or  g?.V  times  the 
r-i'.im  to  be  insured.  Now,  if  $20709  is  .97^  times  the  required  sum, 
once  that  sum  is  $20709-^ .97!,  or  §21240,  Ans.  (Art.  337.)  Hence,  the 

EULE. — Divide  the  value  of  the  property  by  i  minus  the  rate. 

NOTE. — This  and  the  preceding  problem  cover  the  ordinary  cases 
of  Fire  and  Marine  Insurance.  Should  other  problems  occur,  they 
may  be  solved  like  the  corresponding  problems  in  Percentage 


How  find  the  premium,  when  the  sum  insured  and  the  rate  are  given  f 


294  INSURANCE. 

11.  A  merchant  sent  a  cargo  of  goods  worth  $15275  to 
Canton:  what  sum  must  he  get  insured  at  3%,  that  he 
may  suffer  no  loss,  if  the  ship  is  wrecked  ?  ^h.9) 

12.  A  house  and  furniture  are  worth  $27250 :  what  sum 
must  be  insured  at  2%  to  cover  the  property  and  premium? 

13.  What  sum  must  be  insured  at  *>%  to  cover   the 
premium,  with  a  vessel  and  cargo  worth  $35250? 

LIFE     INSURANCE. 

445.  Life  Insurance  Policies  are  of  different  kinds,  and 
the  premium  varies  according  to  the  expectation  of  life. 

ist.  Life  Policies,  which  are  payable  at  the  death  of  the  party 
named  in  the  policy,  the  annual  premium  continuing  through  life. 

2d.  Life  Policies,  payable  at  the  death  of  the  insured,  the 
annual  premium  ceasing  at  a  given  age. 

3d.  Term  Policies,  payable  at  the  death  of  the  insured,  if  he 
dies  during  a  given  term  of  years,  the  annual  premium  continuing 
till  the  policy  expires. 

4th.  Endowment  Policies,  payable  to  the  insured  at  a  given 
age,  or  to  his  heirs  if  he  dies  before  that  age,  the  annual  premium 
continuing  till  the  policy  expires. 

NOTE. — The  expectation  of  life  is  the  average  duration  of  the  life 
of  individuals  after  any  specified  age. 

1.  What  premium  must  a  man,  at  the  age  of  25,  pay 
annually  for  a  life  policy  of  $5000,  at  ^\%  ? 

ANALYSIS. — <$%  =.045,  and  $5000  x  . 045=1225,  Ana.     (Art.  443.) 

2.  What  is  the  annual  premium  for  a  life  policy  of 
£2500,  at  5$  ? 

JL- 3.  A  man  at  the  age  of  35  years  effected  a  life  insurance 
of  $7500  for  10  yrs.,  at  z\%\  what  was  the  amt.  of  premium? 

4.  A  man  65  years  old  negotiated  a  life  insurance  of 
$8000  for  5  years,  at  1 2%% :  what  was  the  amt.  of  premium  ? 

5.  At  the  age  of  30  years,  a  man  got  his  life  insured  for 
'  $75000,  at  4^  per  annum,  and  lived  to  the  age  of  70 :  which 

Was  the  greater,  the  sum  he  paid,  or  the  sum  received? 

444.  How  find  what  sum  must  be  insured  to  cover  the  property  and  preminm  ? 


TAXES. 

446.  A  Tax  is  a  sum  assessed  upon  the  person  or 
property  of  a  citizen,  for  public  purposes. 

A  Property  Tax  is  one  assessed  upon  property. 

A  Personal  Tax  is  one  assessed  upon  the  person, 
and  is  often  called  a  poll  or  capitation  tax. 

NOTE. — The  term  poll  is  from  the  German  polle,  the  head ;  capita- 
tion, from  the  Latin  caput,  the  head. 

447.  Property  is  of  two  kinds,  real  and  personal. 
Heal  Property  is  that  which  is  fixed;  as,  lands, 

houses,  etc.     It  is  often  called  real  estate. 

Personal  Property  is  that  which  is  movable;  as, 
money,  stocks,  etc. 

448.  An  A.ssessor  is  a  person  appointed  to  appraise 
property,  for  the  purpose  of  taxation. 

449.  An  Inventory  is  a  list  of  taxable  property, 
with  its  estimated  value. 

450.  Property  taxes  are  computed  at  a  certain  per  cent  on  the 
valuation  of  the  property  to  be  assessed.     That  is, 

The  valuation  of  the  property  is  the  base ;  the  sum  to  be  raised, 
the  percentage ;  the  per  cent  or  tax  on  $i,  the  rate ;  the  sum  collected 
minus  the  commission,  the  net  proceeds. 

Poll  tuxes  are  specific  sums  upon  those  not  exempt  by  law,  without 
r<;gard  to  property. 

451.    To  assess  a  Property  Tax,  the  Valuation   and  the 
Sum  to  be  raised  being  given. 

i.  A  tax  of  $6250  was  levied  upon  a  corporation  of  6  per- 
sons; A's  property  was  appraised  at  $30000;  B's,  837850; 
C's,  $40150 ;  D's,  $50000 ;  E's,  $55000 ;  F's,  $37000.  What 
was  the  rate  of  the  tax,  and  what  each  man's  share  ? 

446.  What  is  n  tax  ?  A  property  tax  ?  A  personal  tax  ?  Note.  What  is  a  per- 
sonal tax  called  ?  447.  Of  how  many  kinds  is  property '!  What  is  real  property  ? 
l'i!!vonal?  44X.  Whut  is  an  assessor?  449.  An  inventory?  450.  How  are  prop- 
erty tuxes  computed?  How  are  poll  taxes  levied?  451.  How  is  a  property  tax 
assessed,  when  the  valuation  and  the  sum  to  he  raised  are  given? 


396 


TAXES. 


ANALYSIS. — The  valuation=$30ooo  +  $37850  +  $40150  +  $50000  + 
$55000 +  $37000= $250000.  The  valuation  $250000,  is  the  base; 
and  the  tax  $6250,  the  percentage.  Therefore  $625o-i-$25oooo=.O25, 
or  z\fo,  the  rate  required. 

Again,  since  A's  valuation  was  $30000,  and  the  rate  z\%,  his  tax 
must  have  been  "2.\c/o  of  $30000;  and  $30000  x  .025  =  $750.00.  Tho 
tax  of  the  others  may  be  found  in  like  manner.  Hence,  the 

EULE. — I.  Make  an  inventory  of  all  the  taxable  property. 

II.  Divide  the  sum  to  be  raised  by  the  amount  of  the 
inventory,  and  the  quotient  will  be  the  rate. 

III.  Multiply  the  valuation  of  each  man's  property  by 
the  rate,  and  the  product  will  be  his  tax. 

NOTES. — i.  If  a  poll  tax  is  included,  the  sum  arising  from  the 
polls  must  be  subtracted  from  the  sum  to  be  raised,  before  it  is 
divided  by  the  inventory. 

2.  If  the  tax  is  assessed  on  a  large  number  of  individuals,  the 
operation  will  be  shortened  by  first  finding  the  tax  on  $i,  $2,  $3,  etc., 
to  $9  ;  then  on  $10,  $20,  etc.,  to  $90 ;  then  on  $100,  $200,  etc.,  to  $900, 
etc.,  arranging  the  results  as  in  the  following 

ASSESSORS'    TABLE. 


r* 

$1 

pays 

$.025 

$10 

pay 

$.25 

$100 

pay 

$2.50 

2 

M 

.050 

20 

a 

.50 

2OO 

( 

5.00 

3 

tt 

•075 

30 

te 

•75 

300 

I 

7-5° 

4 

11 

.IOO 

40 

a 

I.OO 

40O 

t 

IO.OO 

5 

It 

.125 

50 

tt 

1-25 

500 

( 

12.50 

6 

It 

.150 

60 

it 

1.50 

1000 

t 

25.00 

7 

ft 

•175 

70 

a 

i-75 

2OOO 

(( 

50.00 

8 

it 

.200' 

80 

tt 

2.OO 

3000 

tt 

75.00 

9 

tt 

.225 

9° 

<t 

2.25 

4OOO 

ft 

100.00 

2.  B's  valuation—  $37850  —  $30000  +  $7000  +  $800  +  $50, 

By  the  table  the  tax  on  $30000=8750.00 

"  "  7000=   175.00 

"  "  800=     20.00 

5°=  5 


Therefore,  we  have  B's  tax,        =$946.25 


Note.  If  a  poll  tax  is  included,  how  proceed  ? 


TAXES.  297 

3.  Required  C,  D,  E,  and  F's  taxes,  both  by  the  rule 
and  the  table. 

4.  A  tax  levied  on  a  certain  township  was  $16020;  the 
valuation  of  its  taxable  property  was  $784750,  and  the 
number  of  polls  assessed  at  $1.25  was  260.     What  was  the 
'•  ue  of  tax  ;  and  what  was  A's  tax,  who  paid  for  3  polls, 
the  valuation  of  his  property  being  $7800  ? 

5.  The  State  levied  a  tax  of  $165945  upon  a  certain  city 
which  contained  1260  polls  assessed  75  cents  each,  an  in- 
ventory of  $542.7600  real,  and  §72400  personal  property. 
What  was  G's  tax,  whose  property  was  assessed  at  $15000  ? 

6.  What  was  H's  tax,  whose  inventory  was  $10250,  and 
3  polls  ? 

7.  A  district  school-house  cost  $2500,  and  the  valuation 
of  the  property  of  the  district  is  $50000  :  what  is  the  rate, 
and  what  A's  tax,  whose  property  is  valued  at  $3400  ? 

452.    To  find  the  Amount  to   be  assessed,  to  raise  a  net 
sum,  and  pay  the  Commission  for  collecting  it. 

8.  A  certain  city  required  $47500  to  pay  expenses:  what 
amount  must  be  assessed  in  order  to  cover  the  expenses, 
and  the  commission  of  $%  for  collecting  the  tax  ? 

ANALYSIS.  —  At  5$  for  collection,  $i  assessment  yields  $.95; 
therefore,  to  obtain  $47500  net,  requires  as  many  dollars  assessment 
as  $.95  are  contained  times  in  $47500;  and  $475OO-e-.95  (1  —  5^)  = 
$50000,  the  sum  required  to  be  assessed.  (Art.  341.)  Heiice,  the 

"RULE.  —  Divide  the  net  sum  by  i  minus  the  rate;  the 
quotient  will  be  the  amount  to  be  assessed.  (Art.  348.) 

9.  What  sum  must  be  assessed  to  raise  a  net  amount  of 
^3500,  and  pay  the  commission  for  collecting,  at  4%? 

10.  What  sum  must  be  assessed  to  raise  a  net  sum  of 
$5260,  and  pay  for  the  collection,  at  $\%  commission  ? 

11.  What  sum  must  be  assessed  to  raise  a  net  amount 
^0500,  and  pay  the  commission  for  collecting,  at 


452.  How  find  the  amount  to  be  assessed,  to  cover  the  sum  to  be  raised  and 
the  commission  for  collection  ? 


DUTIES. 

453.  Duties  are  sums  paid  on  imported  goods,  and 
are  often  called  customs. 

454.  A  Custom  House  is  a  building  where  duties  are  re- 
ceived, ships  entered,  cleared,  etc. 

455.  A  Fort  of  Entry  is  one  where  there  is  a  Custom  House. 
The  Collector  of  a  Port  is  an  officer  who  receives  the  duties, 

has  the  charge  of  the  Custom  House,  etc. 

456.  A  Tariff  is  a  list  of  articles  subject  to  duty,  stating  the 
rate,  or  the  sum  to  be  collected  on  each. 

457.  An  Invoice  is  a  list  of  merchandise,  with  the  cost  of  the 
several  articles  in  the  country  from  which  they  are  imported. 

458.  Duties  are  of  two  kinds,  specific  and  ad  valorem. 

459.  A  Specific  Duty  is  a  fixed  sum  imposed  on 
each  article,  ton,  yard,  etc.,  without  regard  to  its  cost. 

An  Ad  Valorem  Duty  is  a  certain  per  cent  on  the 
value  of  goods  in  the  country  from  which  they  are 
imported. 

NOTE. — The  term  ad  valorem  is  from  the  Latin  ad  and  valorem, 
according  to  value. 

460.  Before  calculating  duties,  certain  allowances  are  made, 
called  tare,  tret  or  draft,  leakage,  and  breakage. 

Tare  is  an  allowance  for  the  weight  of  the  box,  bag,  cask,  etc., 
containing  the  goods. 

Tret  is  an  allowance  in  the  weight  or  measure  of  goods  for  waste 
or  refuse  matter,  and  is  often  called  draft. 

Leakage  is  an  allowance  on  liquors  in  casks. 

Breakage  is  an  allowance  on  liquors  in  bottles. 

NOTES. — i.  Tare  is  calculated  either  at  the  rate  specified  in  the 
invoice,  or  at  rates  established  by  Act  of  Congress. 

2.  Leakage  is  commonly  determined  by  gauging  the  casks,  and 
Breakage  by  counting. 

3.  In  making  these  allowances,  if  the  fraction  is  lees  than  \  it  is 
rejected,  if  \  or  more,  i  is  added. 

453.  What  are  duties ?  454.  A  custom  house?  455.  What  is  a  port  of  entry T 
A  collector  ?  456.  A  tariff?  457.  What  is  an  invoice  ?  458.  Of  how  many 
kinds  are  duties?  459.  A  specific  duty?  An  ad  valorem?  460.  What  ia  tareT 
Tret  ?  Leakage  ?  Breakage  ? 


DUTIES.  299 

PROBLEM     I. 

461.  To  calculate  Specific  Duties,  the  quantity  of  goods, 
and  the  Sum  levied  on  each  article  being  given. 

1.  What  is  the  specific  duty  on  12  casks  of  brandy,  each 
containing  40  gal.,  at  $i-£  per  gal.,  allowing  2%  for  leakage  ? 

ANALYSIS. — 12  casks  of  40  gallons  each,  =480.0  gal. 

The  leakage  at  z%  on  480  gallons,  =     9.6  gal. 

The  remainder,  or  quantity  taxed,          =  47°-4  gal. 
Now  $1.50  x  470.4=$ 705.60,  the  duty  required.    Hence,  the 

RULE. — Deduct  the  legal  allowance  for  tare,  tret,  etc., 
from  the  goods,  and  multiply  the  remainder  by  the  sum 
levied  on  each  article. 

2.  What  is  the  duty,  at  $1.25  a  yard,  on  65  pieces  of 
brocade  silk,  each  containing  50  yards  ? 

3.  What  is  the  duty  on  87  hhds.  of  molasses,  at  20  cts. 
per  gallon,  the  leakage  being  3%  ? 

4.  What  is  the  duty,  at  6  cts.  a  pound,  on  500  bags  of 
coffee,  each  weighing  68  Ibs.,  the  tare  being  2%  ? 

PROBLEM    II. 

462.  To  calculate  Ail    Valorem,  Duties,  the   Cost  of  the 

goods  and  the  Rate  being  given. 

5.  What  is  the  ad  valorem  duty,  at  25$,  on  a  quantity 
of  silks  invoiced  at  $3500  !J 

ANALYSIS. — Since  the  duty  is  25$  ad  valorem,  $35°° 

it  is  .25  times  the  invoice;  and  $3500 x  .25  =  1875.  .25 

Hence,  the  Ans,  $875.00 

RULE. — Multiply  the  cost  of  the  goods  by  the  given  rate, 
expressed  decimally. 

6.  What  is  the  duty,  at  ZZ\%  a^  valorem,  on  1575  yds. 
of  carpeting  invoiced  at  $1.80  per  yard? 

7.  What  is  fhe  ad  valorem  duty,  at  40%,  on  no  chests 
of  tea,  each  containing  67  Ibs.,  and  invoiced  at  90  cts.  a 
pound,  the  tare  being  9  Ibs.  a  chest  ? 

461.  How  calculate  specific  duties  when  the  quantity  of  goods  and  the  sum 
evied  on  eacli  article  are  given  ?  462.  How  ad  valorem  ? 


INTERNAL    REVENUE. 

463.  Internal  Revenue  is   the   income   of  the 
Government  from  Excise  Duties,  Stamp  Duties,  Licenses, 
Special  Taxes,  Income  Taxes,  etc. 

464.  Excise  Duties  are  taxes  upon  certain  home  productions,, 
and  are  computed  at  a  given  per  cent  on  their  value. 

465.  Stamp  Duties  are  taxes  upon  written  instruments ;  as, 
notes,  drafts,  contracts,  legal  documents,  patent  medicines,  etc. 

466.  A  License  Tax  is  the  sum  paid  for  permission  to  pursue 
certain  avocations. 

467.  Special   Taxes  are  fixed  sums  assessed  upon  certain 
articles  of  luxury;  as,  carriages,  billiard  tables,  gold  watches,  etc. 

Income  Taxes  are  those  levied  upon  annual  incomes. 

NOTE.— In  determining  income  taxes,  certain  deductions  are  made 
for  house  rent,  National  and  State  taxes,  losses,  etc. 

468.  To  compute  Income  Taxes,  the  Rate  being  given. 

1.  What  is  a  man's  tax  whose  income  is  $5675;   the 
rate  being  5^',  the  deductions  $1000  for  house  rent,  $350 
national  tax,  and  -Si  100  for  losses  ? 

ANALYSIS. — Total  deductions  are  $1000  + $350  + $iioo=$245o; 
and  $5675— $245O=$3225  taxable  income.  Now  $%  of  $3225  = 
3225  x  .05 =$161.25,  *ue  tax  required^  Hence,  the 

KULE. — From  the  income  subtract  the  total  deductions, 
and  multiply  the  remainder  by  the  rate. 

NOTE. — If  there  are  special  taxes  on  articles  of  luxury,  as  carriages, 
etc.,  they  must  be  added  to  the  tax  on  the  income. 

2.  What  was  A's  revenue  tax  for  1869,  at  5% ;  his  income 
being  $4750;  his  losses  $1185,  and  exemptions  £1200  ? 

3.  In  1870,  A  had  $10500  income;  35  oz.  taxable  plate 
at  5  cts.,  i  watch  $2,  and  i  carriage  $2 :  what  was  his  tax 
at  5%,  allowing  $2100  exemption? 

463.  What  is  internal  revenue?  464.  Excise  duties?  463.  Stamp  duties? 
466.  A  license  tax?  467.  Special  taxes?  Income  taxes?  468.  How  compute 
Income  taxes,  when  the  rate  is  given  ? 


EQUATION    OF    PAYMENTS. 

469.  Equation  of  Payments  is  finding  the  average 
time  for  payment  of  two  or  more  sums  due  at  different  times. 

The  average  time  sought  is  often  called  the  mean,  or 
ciuated  time. 

470.  Equation  of  Payments  embraces  two  classes  of  examples. 
ist.  Those  in  which  the  items  or  bills  have  the  same  date,  but 

different  lengths  of  credit.     2d.  Those  in  which  they  have  different 
dates,  and  the  same  or  different  lengths  of  credit. 

PROBLEM    I. 

471.  To  find  the  Average  Time,  when  the  items  have  the 
same  date,  but  different  lengths  of  credit. 

i.  Bought  Oct.  3d,  1870,  goods  amounting  to  the  fol- 
lowing sums:  $50  payable  in  4  m.,  $70  in  6  m.,  and  $80  in 
8  m. :  what  is  the  average  time  at  which  the  whole  may  be 
paid,  without  loss  to  either  party? 

ANALYSIS. — The  int.  on  $50  for  4  m.=the  int.  $^0  X4=2OO 
on  $i  for  50  times  4  m.  or  200  m.  Again  the  int.  -o  x  5  —  420 
on  $70  for  6  m.=.the  int.  on  $i  for  70  times  6  m.,  gQ  x  5  =  640 
or  420  in.  Finally,  the  int.  on  $80  for  8  m.=the  — . — — 

int.  of  $i  for  80  times  8  m.,  or  640  m.    Now  200 
m.  +  42o  m. 4-640  m.  =  i26o  m. ;    therefore  I  am      Ans.  6-j^  m. 
entitled  to  the  use  of  $i  for  1260  m.     But  the  sum 
of  the  debts  is  $50  +  $70  +  $80= $200.     Now  as  I  am  entitled  to  the 
use  of  $i  for  1260  m.,  1  must  be  entitled  to  the  use  of  $200  for  ^  part 
of  1260  in.,  and  i26o-^2oo=6,3o  m.,  or  6  m.  9  d.,  the  average  time  re- 
quired.    Hence,  the 

RULE. — Multiply  each  item  by  its  length  of  credit,  and 
ill  ride  the  sum  of  the  products  by  the  sum  of  the  items. 
The  quotient  will  be  the  average  time. 

NOTES. — i.  When  the  date  of  the  payment  is  required,  add  the 
average  time  to  the  date  of  the  transaction.  Thus,  in  the  preceding 
Ex.,  the  date  of  payt.  is  Oct.  3d +  6  m.  9  d.,  or  April  i2th,  1871. 

469.  What  is  equation  of  payments  ?  Note.  What  is  the  average  time  called  T 
470.  How  many  classes  of  examples  in  Equation  of  Payments?  471.  How  find 
the  average  time,  when  the  Items  have  the  same  date,  but  different  credit* 


302  EQUATION     OF     PAYMENTS. 

2.  This  rule  is  applicable  to  notes  as  well  as  accounts.    It  ia 
founded  upon  the  supposition  that  bank  discount  is  the  same  aa 
simple  interest.    Though  not  strictly  accurate,  it  is  in  general  use. 

3.  If  one  item  is  cash,  it  has  no  time,  and  no  product;  but  in  find- 
ing the  sum  of  items,  this  must  be  added  with  the  others. 

4.  In  the  answer,  a  fraction  less  than  £  day,  is  rejected ;  if  |  day 
or  more,  i  day  is  added. 

2.  Bought  a  house  June  2oth,  1 870,  for  $3000,  and  agreed 
to  pay  \  dpwn;  |  in  6  m.,  the  balance  in  12  m. :  at  what 
date  may  the  whole  be  equitably  paid  ? 

3.  A  owes  B  $7ocy  payable  in  4  mos. ;  $500,  in  6  mos. ; 
$800,  in  10  mos.;  and  $1000  in  12  mos.:  in  what  time 
may  the  whole  be  justly  paid? 

4.  Bought  a  bill  of  goods  March  roth,  1868,  amounting 
to  $2500;  and  agreed  to  pay  $500  cash,  $750  in  10  days, 
$600  in  20  days,  $400  in  30  days,  and  $250  in  40  days. 
At  what  date  may  I  equitably  pay  the  whole  ? 

5.  A  jobber  sold  me  on  the  ist  of  March  $12000  worth 
of  goods,  to  be  paid  for  as  follows :  J  cash,  £  in  2  mos.,  £  in 
4  mos.,  and  the  remaining  -£  in  6  mos.    When  may  1  pay 
the  whole  in  equity  ? 

PROBLEM    II. 

472.    To   find   the   Average    Titnc,   when   the    items   have 
different  dates,  and  the  same  or  different  lengths  of  credit. 

6.  Bought  the  following  bills  of  goods:   March  loth, 
1870,  $500  on  2  m. ;  April  4th,  $800  on  4  m. ;  June  i5th, 
1 1  ooo  on  3  m.     What  is  the  average  time? 

ANALYSIS.— We  first  find  Mayio,  oo  x  $500=  oooo  d. 
when  the  items  are  due  by  A  ^  86  x  $8oo-  68800  (1. 
adding  the  time  of  credit  to  g  t  J2g  x  $IOOO  =  128000  (1. 
the  date  of  each,  and  for  con-  -r —  -  . — 7-5- 

venience  place   these  dates 

in  a  column.    Taking  the  Average  time,  &5-11  d. 

earliest  date  on  which  cither 

item  matures  as  a  standard,  we  find  the  number  of  days  from  this 
standard  date  to  the  maturity  of  the  other  items,  and  place  them  on 
the  right,  with  the  sign  (  x )  and  the  items  opposite. 


EQUATION     OF     PAYMENTS.  303 

Multiplying  each  item  and  its  number  of  days  together,  the  sum 
of  the  products  shows  that  the  interest  on  the  several  items  is  equal 
to  the  interest  of  $i  for  196800  days. 

Now  if  it  takes  196800  days  for  $i  to  gain  a  certain  sum,  it  will 
take  $2300,  jAiu  of  196800  d.  to  gain  this  sum  ;  and  196800-^-2300= 
85^,  or  86  days  from  the  assumed  date.  Now  May  10  +  86  d.=Aug. 
4th,  the  date  when  the  amt.  is  equitably  due.  Hence,  the 

RULE. — I.  Find  the  date  when  the  several  items  become 
due,  and  set  them  in  a  column. 

Take  the  earliest  of  these  dates  as  a  standard,  and  set  the 
number  of  days  from  this  date  to  the  maturity  of  the  other 
items  in  another  column  on  the  right,  with  the  items  opposite. 

II.  Multiply  each  item  by  its  number  of  days,  and  divide 
the  sum  of  the  products  by  the  sum  of  the  items.  The 
quotient  will  be  the  average  time  of  credit. 

NOTES. — i.  Add  the  average  time  to  the  standard  date,  and  the 
result  will  be  the  equitable  date  of  payment. 

2.  The  latest  date  on  which  either  item  falls  due  may  also  be 
taken  as  the  standard ;  and  having  found  the  average  time,  subtract 
it  from  this  date  ;  the  result  will  be  the  date  of  payment. 

3.  The  date  at  which  each  item  becomes  due,  is  readily  found  by 
adding  its  time  of  credit  to  the  date  of  the  transaction. 

7.  A  bought  goods  as  follows:  Apr.  loth,  $310  on  6  m. ; 
May  zist,  $468  on  2  m. ;  June  ist,  $520  on  4  m. ;  July  8th, 
6750  on  3  m. :  what  is  the  average  time,  and  at  what  date 
may  the  whole  debt  be  equitably  discharged  ? 

SUGGESTION. — The  second  item  matures  earliest;  this  date  is 
therefore  the  standard.  Ans.  Av.  time  59  d. ;  date  Sept.  i8th. 

8.  Sold  the  following  bills  on  6  months  credit:  Jan. 
i5th,  $210;  Feb.  nth,  $167;  March  7^1,1320.25;  April 
2d,  8500.10:  when  may  the  whole  be  paid  at  one  time? 

9.  Bought  June  5th,  on  4  m.,  groceries  for  $125  ;  June 
nst,  $230.45;  July  1 2th,  $267;  Aug.  2d,  $860.80:  what 
is  the  amt.  and  time  of  a  note  to  cover  the  whole  ? 

472.  How  find  the  average  tiaie,  when  the  items  have  different  dates,  and  the 
same  or  different  credits  ?  Note.  How  find  the  date  of  the  payment  ?  How  find 
the  date  at  which  each  item  becomes  due  ? 


304 


AVERAGING     ACCOUNTS. 


AVERAGING     ACCOUNTS. 

473.  An  Account  is  a  record  of  the  items  of  debit 
and.  credit  in  business  transactions. 

NOTE. — The  term  debit  is  from  the  Latin  detritus,  owed. 

474.  A  Merchandise  Balance  is  the  difference 
between  the  debits  and  credits  of  an  account. 

A  Cash  Balance  is  the  difference  between  the  debits 
and  credits,  with  the  interest  due  on  each. 

NOTES. — i.  Bills  of  goods  sold  on  time  are  entitled  to  interest 
after  they  become  due  ;  and  payments  made  before  they  are  due  PTC 
also  entitled  to  interest. 

475.  To  find  the  Average  Time  for  paying  the  balance  of 
an  Account  which  has  both  debits  and  credits. 

i.  Find  the  cash  Bal.  of  the  following  Acct.,  and  when  due. 
Dr.  WM.  GORDON  in  Acct.  with  JOHN  KANDOLPH.  Cr. 


1870. 

Feb.  10. 
May  ii. 

July  26. 

ForMdse.,  4  m. 

::   ::  \: 

$450.00 
500.00 
360.00 

1870. 

Mar.  20. 
July    9. 
Sept.  15. 

By  Sundries,  3  m. 
"   Draft,      60  d. 
"   Cash, 

$325-00 
150.00 
400.00 

ANALYSIS.  —  Setting 
down  the  date  when  each 
item  is  due.take  June  loth, 
the  earliest  of  these  dates, 
as  the  standard,  find  the 
number  of  days  from  this 
standard  to  the  maturity 
of  each  item  on  both  sides, 
and  place  it  on  the  right 
of  its  date  with  the  sign 
(  x  ),  then  the  items,  add- 
ing 3  days  grace  to  the 
time  of  the  draft. 


Debits. 

June  10,  oo  d.  x  $450  =  ooooo  d. 
Aug.  n,  62  d.  x  500  =  31000  d. 
Sept.  26,  108  d.  x  360  =  38880  d. 

Amt.  debits,  81310,  Int.  69880  d. 

Credit*. 

June  20,    10  d.  x$325  =    3250  d. 

Sept.  10,    92.  d.  x    150  =  13800  d. 

"      J5>    97  d-  x   400  =  38800  d. 

Amt.  credits,  $875,1^.55850  d. 

Cash  bal.  $435,   "   14030  d. 

$435)14030  d.  =  32  d.+. 
Ans.  Bal.  $435,  due  July  12,  '70. 


473.  What  is  an  account?     Note.  What  is  the  meaning  of  the  term  debit  J 
.  What  is  a  merchandise  balance  f    A  cash  balance  ? 


AVERAGING     ACCOUNTS.  305 

Multiplying  each  item  on  both  sides  by  its  number  of  days, 
the  int.  on  the  debits  is  equal  to  the  int.  of  $i  for  69880  days, 
and  the  int.  on  the  credits  is  equal  to  the  int.  of  $i  for  55850  days. 
(Art.  474.)  The  balance  of  int.  on  the  Dr.  side  is  69880  d.— 55850  d. 
=  14030  d. ;  that  is  to  the  int.  of  $i  for  14030  d.  The  balance  of 
items  on  the  Dr.  side  is  $1310— $875=$435. 

Now  if  it  takes  14030  days  for  $i  to  gain  a  certain  sum,  it  will 
take  $435,  4 h-  of  14030 d.  and  14030-^-435=3215,  or  32  d.  But  the 
assumed  standard  June  ioth  +  32  d.=July  i2th.  Therefore  the 
balance  due  Randolph  the  creditor  is  $435,  payable  July  I2th,  1870. 

If  the  greater  sum  of  items  and  the  greater  sum  of  products  were 
»n  opposite  sides,  it  would  be  necessary  to  subtract  the  average  time 
from  the  assumed  date.  Hence,  the 

RULE. — I.  Set  down  the  date  when  each  item  of  debit  and 
credit  is  due;  and  assuming  the  earliest  of  these  dates  as  a 
standard,  write  the  number  of  days  from  this  standard 
date  to  the  maturity  of  the  respective  items,  on  the  right, 
with  the  sign  (  x )  and  the  items  themselves  opposite. 

II.  Multiply  each  item  by  its  number  of  days,  and  divide 
the  difference  between  the  sums  of  products  by  that  between 
the  sums  of  items ;  the  quotient  will  be  the  average  time. 

III.  If  the  greater  sum  of  items  and  the  greater  sum  of 
products  are  both  on  the  same  side,  add  the  average  time  to 
the  assumed  date;  if  on  opposite  sides,  subtract  it;  and 
the  result  will  be  the  date  when  the  balance  of  the  account  is 
equitably  due. 

NOTES. — i.  The  average  time  may  be  such  as  to  extend  to  a  date 
either  earlier  or  later  than  that  of  any  of  the  items.  (Ex.  2.) 

2.  In  finding  the  maturity  of  notes  and  drafts,  3  days  grace  should 
be  added  to  the  specified  time  of  payment. 

3.  In  finding  the  extension  to  which  the  balance  of  a  debt  is 
entitled,  when  partial  payments  are  made  before  it  is  due, 

Multiply  each  payment  by  the  time  from  its  date  to  the  maturity 
of  the  debt,  and  divide  the  sum  of  the  products  by  the  balance 
remaining  unpaid. 

4.  When  no  time  of  credit  is  mentioned,  the  transaction  is  under- 
stood to  be  for  cash,  and  its  payment  due  at  once. 

475.  How  find  the  average  time  for  paying  the  balance,  when  there  are  both 
debits  and  credits  T 


306 


AVERAGING     ACCOUNTS. 


2.  Find  the  balance  of  the  following  Acct.,  and  when  due. 
Dr.  A.  B.  in  account  with  C.  D.  Cr. 


1860. 

1860. 

Aug.  ii. 

For  Mdse., 

$160.00 

Sept.  2. 

By  Sundries, 

$75-oo 

Sept.    5. 

"        " 

240.00 

Oct.  10. 

"   your  Note  on  30  d., 

100.00 

Oct.   20. 

"   i  Horse, 

175-00 

Nov.  i. 

"    Cash 

110.00 

Dr. 


OPERATION. 


Cr. 


Due. 

Days. 

Items. 

Products. 

Due. 

Days. 

Items. 

Products. 

Aug.  II. 

0 

$160 

0000 

Sept.    2. 

22 

$75 

1650 

Sept.    5. 

25 

240 

6000 

Nov.  )£ 

93 

100 

9300 

Ot.    20. 

70 

175 

12250 

Nov.    i. 

82 

no 

9020 

575 

18250 

285 

19970 

^?1 

1720-4-290=6  d.  (nearly),  av.  tune. 

18250 

Bal=$2go 

Aug.  ii—  6d.=Aug.  5,bal.isdue. 

1720 

SUGGESTION. — In  this  example,  the  greater  sum  of  items  and  the 
greater  sum  of  products  are  on  opposite  bides ;  hence,  the  average 
time  must  be  subtracted  from  the  standard  date. 

3.  A  man  bought  a  cottage  for  $2 7 50, payable  in  i  year; 
in  3  mos.  he  paid  $500,  and  3  mos.  later  $750:  to  what 
extension  is  he  entitled  on  the  balance  ? 

SOLUTION. — The  sum  of  the  product  is  9000 ;  and  the  balance  of 
the  debt  $1500.  Now  9OOO-7-i5oo=:6.  Ans.  6  mcs.  after  the  maturity 
of  the  debt.  (Note  3.) 

4.  A  merchant  sold  a  bill  of  $4220  worth  of  goods  on 
8  mos.;  2  months  after  the  customer  paid  him  $720,  one 
month  later  $850,  and  2  months  later  $1000:  how  long 
should  the  balance  in  equity  remain  unpaid  ? 

5.  "What  is  the  balance  of  the  following  account,  and 
when  is  it  due  ? 

Dr.    HENRY  SWIFT  in  Acct.  with  HOMER  MORGAN.    Cr. 


\   1865. 

1865. 

•March  10. 

For  Sundries, 

$2  SO 

!  April     i. 

By  Bal.  of  Acct., 

Sno 

JApril    15. 

"    Flour  on  60  d., 

420 

iMav     21. 

"   Dft,  onsod.. 

300 

'June    20. 

"    Mdse.  on  30  d., 

600 

July      i. 

"   Cash, 

560' 

RATIO. 

476.  Ratio  is  the  relation  which  one  number  has  to 
another  with  respect  to  magnitude. 

The  Terms  of  a  ratio  are  the  numbers  compared. 
They  are  often  called  a  couplet. 

477.  Ratio  is  commonly  expressed  by  a  colon  (:)  placed  between 
the  two  numbers  compared.    Thus,  the  ratio  of  6  to  3  is  written  6  :^, 

478.  The  first  term  is  called  the  antecedent,  the  second  the 
consequent. 

The  comparison  is  made  by  considering  what  multiple  or  part  the 
antecedent  is  of  the  consequent. 

NOTES. — i.  The  sign  of  ratio  (:)  is  derived  from  the  sign  of 
division  (-=-),  the  horizontal  line  being  dropped. 

2.  The  terms  are  so  called  from  the  order  of  their  position.  They 
must  be  of  the  same  kind  or  denomination;  otherwise  they  cannot 
be  compared. 

479.  Ratio  is  measured  by  &  fraction,  the  numerator  of  which  is 
the  antecedent,  and  the  denominator  the  consequent;  or  what  is  the 
same  thing,  by  dividing  the  antecedent  by  the  consequent. 

480.  Th0  value  of  a  ratio  is  the  value  of  the  fraction  by  which 
it  is  measured.     Thus,  comparing  6  with  2,  we  say  the  ratio  of  6 :  2 
is  5,  or  3.    That  is,  the  former  has  a  magnitude  which  contains  tho 
latter  3  times ;  therefore  the  value  of  the  ratio  6 : 2  is  3. 

481.  A  Simple  Hatio  is  one  which  has  but  two 

terms ;  as,  8 : 4. 

A  Compound  Hatio  is  the  product  of  two  or  more 
simple  ratios. 

Thus :  4 :  2  \  are  simple  But  4x9:2x3 

9:3)      ratios.  is  a  compound  ratio. 

NOTE. — The  nature  of  compound  ratios  is  the  same  as  that  of 
simple  ratios.  They  are  so  called  to  denote  their  origin,  and  are 
usually  expressed  by  writing  the  corresponding  terms  of  the  simple 

ratios  one  under  another,  as  above. 

\ 

476.  What  is  ratio?  The  numbers  compared  called ?  477.  How  is  ratio  com- 
monly expressed  ?  The  first  term  called  ?  The  second  ?  Note.  Why  ?  479.  How 
is  ratio  measured  ?  The  value  of  a  ratio  ?  481.  Simple  ratio?  Compound? 


308  RATIO. 

482.  Ratio  is  also  distinguished  as  direct  and  inverse  or  reciprocal. 

A  direct  ratio  is  one  which  arises  from  dividing  the  antecedent  by 
the  consequent. 

An  inverse  ratio  is  one  which  arises  from  dividing  the  consequent 
by  the  antecedent,  and  is  the  same  as  the  ratio  of  the  reciprocals 
of  the  two  numbers  compared.  (Art.  106.)  Thus,  the  direct  ratio 
of  4  to  i2=-,*2,  or  £  ;  the  inverse  ratio  of  4  to  I2=i4i,  or  3.  It  is  the 
same  as  the  ratio  of  their  reciprocals,  \  to  -f^.  , 

483.  1'he  ratio  between  two  fractions  having  a  common  denomi- 
nator is  the  same  as  the  ratio  of  their  numeratort.    Thus,  the  ratio 
of  § :  3  is  the  same  as  6 : 3. 

In  finding  the  ratio  of  two  fractions  which  have  different  denomi- 
nators, they  should  be  reduced  to  a  common  denominator;  then  take 
the  ratio  of  their  numerators.  (Art.  153.) 

484.  In  finding  the  ratio  between  two  compound  numbers,  they 
must  be  reduced  to  the  same  denomination. 

NOTE. — Finding  the  ratio  between  two  numbers  is  the  same  in 
principle  as  finding  what  part  one  is  of  the  other,  the  number 
denoting  the  part  being  the  antecedent.  Thus,  2  is  J  of  4=^ ;  and 
the  ratio  of  2  to  4  is  f =|.  (Art.  173.) 

485.  Since   ratios  are  measured  by  fractions  whose 
numerators  are    the  antecedents,  and  denominators  the 
consequents,  it  follows  that  operations  have  the  same  effect 
upon  the  terms  of  a  ratio  as  upon  the  terms  of  a  fraction. 
(Art.  144.)     That  is, 

1.  Multiplying  the  antecedent  multiplies  the  ratio;  and 
dividing  the  antecedent  divides  the  ratio. 

2.  Multiplying  the  consequent   divides  the  ratio;  and 
dividing  the  consequent  multiplies  the  ratio. 

3.  Multiplying  or  .dividing  loth  the  antecedent  and  con- 
sequent by  the  same  number,  does  not  alter  the  value  of  the 
ratio. 

482.  Direct?  Inverse?  483.  The  ratio  of  two  fractions  having  a  common 
denominator  ?  484.  In  finding  the  ratio  of  two  compound  numbers,  what  must 
be  done?  485.  What  effect  do  operations  on  the  terms  of  a  ratio  have?  Multi- 
plying the  antecedent  ?  Dividing  it  ?  Multiplying  the  consequent  ?  Dividing  it  ? 
What  effect  has  multiplying  or  dividing  both  the  antecedent  »nd  consequent  by 
the  same  number  ? 


PBOPORTIOtf.  309 

What  are  the  ratios  of  the  following  couplets : 

1.  12:4.  4.  6 :  24.  7.  £5  :  ios.  6d. 

2.  28:7.  5-  8:40.  8.  10  y.  :  6ft.  3  in. 
3.36:12.           -6.9:51.  9.  25  g. :  2  qt.  i  pt. 

10.  Reduce  the  ratio  of  14  to  35  to  the  lowest  terms? 
SOLUTION. — 14 : 35  equals  ^  ;  and  i*=f,  or  2  :  5.     (Art.  146.) 
Reduce  the  following  ratios  to  the  lowest  terms  ? 

11.  154:28.     -T3.  73:5ii./>         15.  238:1428. 

12.  39:165.         14.  113:1017.  ,     16.  576:1728. 

17.  Reduce  the  ratio  £  :|  to  the  lowest  integral  terms? 
|=iJ.  and  $ =£f.    Now  iJ  :  £J  is  the  same  as  4  :  5.    (Art.  483.) 

18.  Multiply  the  ratio  of  21 :  7  by  4 :  8.  Ans.  84  :  56,  or  i£. 

19.  What  is  the  value  of  5:8x4:  10x7:9? 


PROPORTION. 

486.  Proportion  is  an  equality  of  ratios. 

487.  Every  proportion  must  have  at  least  four  terms  •, 
for,  the  equality  is  between  two  or  more  ratios,  and  each 
ratio  has  tiuo  terms,  an  antecedent  and  a  consequent. 

A  proportion  may,  however,  be  formed  from  three  num- 
bers; for,  one  of  the  numbers  may  be  repeated,  so  as  to 
form  two  terms. 

488.  Proportion  is  denoted  in  two  ways ;  by  a  double 
colon  ( : : ),  and  by  the  sign  of  equality  (  =  ),  placed  between 
the  ratios.     Thus,  each  of  the  expressions  4 :  2  : :  6 : 3,  and 
4 :  2  =  6 :  3  indicates  a  proportion ;  for,  f=|. 

The  former  is  read,  "  4  is  to  2  as  6  to  3,"  or  "  4  is  the 
same  part  of  2,  that  6  is  of  3."  The  latter  is  read,  "  t he- 
ratio  of  4  to  2  equals  the  ratio  of  6  to  3." 

NOTE. — The  sign  (::)  is  derived  from  the  sign  (=),  the  points 
being  the  extremities  of  the  parallel  lines. 

4S6.  What  is  proportion  ?  4S7.  How  many  terms  has  every  proportion  ?  Can 
three  numbers  form  a  proportion  ?  How  ? 


tit  PROPORTION. 

489.  The  four  numbers  which  form  a  proportion,  are 
called  proportionals.    The  first  and  last  arc  the  extremes, 
the  other  two  the  means. 

When  a  proportion  has  but  three  numbers,  the  second 
term  is  called  a  mean  proportional  between  the  other  two. 

490.  If  four  numbers  are  proportional,  the  product  of 
the  extremes  is  equal  to  the  product  of  the  means.     Hence, 

491.  The  relation  of  the  four  terms  of  a  proportion  to 
each  other  is  such,  that  if  any  three  of  them  are  given,  the 
other  or  missing  term  may  be  found. 

492.  To  find  the  Missing  Term  of  a  Proportion,  the  other 
three  Terms  being  given. 

1.  Let  6  be  the  first   term   of  a  proportion,  3  and  10  the   two 
means;   the  other   extreme  equals   3x10-5-6=5;   for,  the  product 
of  the  means=the  product  of  the  extremes;  and  the  product  of  two 
factors  divided  by  one  of  them,  gives  the  other.     (Art.  93.) 

2.  Let  3,  10  and  5  be  the  last  three  terms  of  a  proportion,  the  first 
term  equals  3  x  10-5-5=30-5-5  or  6. 

3.  Let  6  and  5  be  the  extremes  of  a  proportion,  and  3  one  of  the 
means ;  the  other  mean  equals  6  x  5-5-3=30-5-3  or  10. 

4.  If  6  and  5  are  the  extremes,  and  10  one  of  the  means,  the  other 
mean  equals  6  x  5-5-10=30-5-10  or  3.     Hence,  the 

KTJLE. — I.  If  one  of  the  extremes  and  the  two  means  are 
given,  divide  the  product  of  the  means  by  the  given  extreme. 

II.  If  one  of  the  means  and  the  two  extremes  are  given, 
divide  the  product  of  the  extremes  ly  the  given  mean. 

Find  the  missing  term  in  the  following  proportions : 

1.  52:13: :  62  :  — .  5.  4  rods  :  1 1  ft. : :  18  men :  — . 

2.  15:90::  —  '.72.  6.  24yd. :  3  yd. ::  —  :$i2. 

3.  60 :  — : :  100  :  33^.         7.  20  gal. :  — : :  $40  :  $8. 

4.  — :  25  : :  £ :  £.  8.  — :  40  Ib. : :  £2  :  8s. 

489.  What  are  the  four  numbers  forming  a  proportion  called?  491.  If  three 
terms  of  a  proportion  arc  given,  what  is  true  of  the  fourth  ?  If  the  two  means 
and  one  extreme  arc  given,  how  find  the  other  extreme  f  If  the  two  extremes 
and  one  mean  are  given,  how  find  the  other  mean  ? 

-f  i 


•  1MPL1 

493.  Simple  Proportion  is  an  equality  of  two 
simple  ratios. 

Simple  Proportion  is  applied  chiefly  to  the  solu- 
tion of  problems  having  three  terms  given  to  find  &  fourth, 
of  which  the  third  shall  be  the  same  multiple  or  part,  as 
i\\Q  first  is  of  the  second. 

494.  To  solve  Problems  by  Simple  Proportion. 

1.  If  5  baskets  of  peaches  cost  $10,  what  will  3  baskets  cost? 

ANALYSIS. — The  question  assumes  STATEMENT. 

that  3  baskets  can  be  bought  at  the  5  bas. :  3  bas. :  :  $10  rAns. 
same  rate  as  5  baskets ;  therefore  3 

5  bas.  lias  the  same  ratio  to  3  bas.  5)^30 

as  the  cost  of  5  bas.  has  to  the  cost  of  — *7    . 

3  bas.    That  is,  5  bas. :  3  bas. : :  $10 : 

cost  of  3  bas.  We  have  then  the  two  means  and  one  extreme  of  a 
proportion  to  find  the  fourth  term,  or  other  extreme.  (Art.  492.) 
Now  the  product  of  the  means  $10  x  3 =$30 ;  and  $30-1-5  (the  other 
extreme) =$6,  the  cost  of  3  baskets.  Hence,  the 

RULE. — I.  Take  that  number  for  the  third  term,  which  is 
the  same  kind  as  the  answer. 

II.  When  the  answer  is  to  be  larger  than  the  third  term, 
place  the  larger  of  the  other  tivo  numbers  for  the  second 
term;  but  ivhen  less,  place  the  smaller  for  the  second  term, 
and  the  other  for  the  first. 

III.  Multiply  the  second  and  third  terms  together,  and 
divide  the  product  by  the  first;  the  quotient  will  be  the 
fourth  term  or  answer.     (Arts.  490,  491.) 

PROOF. — If  the  product  of  the  first  and  fourth  terms 
equals  that  of  the  second  and  third,  the  answer  is  right. 

NOTES. — i  The  arrangement  of  the  given  terms  in  the  form  of  a 
proportion  is  called  "  Stating  the  question." 

2.  After  stating  the  question,  the  factors  common  to  the  first  and 
second,  or  to  the  first  and  third  terms,  should  be  cancelled. 

493.  Simple  Proportion  ?    4  >4.  How  solve  problems  by  Simple  Proportion  ? 


312  SIMPLE     PBOPORTION. 

3.  If  the  first  and  second  terms  contain  different  denominations, 
they  must  be  reduced  to  the  same.  If  the  third  term  is  a  compound 
number,  it  must  be  reduced  to  the  lowest  denomination  it  contains. 

495.  REASONS. — i.  The  reason  for  placing  that  number  for  the 
third  term,  which  is  the  same  kind  as  the  answer,  and  the  other  two 
numbers  for  the  first  and  second,  is  because  money  has  a  ratio  to 
money,  but  not  to  the  other  two  numbers ;  and  ihe^ther  two  Humbert 
have  a  ratio  to  each  other,  but  not  to  money. 

2.  Of  the  two  like  numbers,  the  smaller  is  taken  for  the  second 
term,  and  the  larye.r  for  the  first,  because  3  baskets  being  less  than 
5  baskets,  will  cost  less ;  consequently,  the  answer  or  fourth  term 
must  be  less  than  the  third,  the  cost  of  5  baskets. 

3.  If  it  were  required  to  find  the  cost  of  a  quantity  greater  than 
that  whose  cost  is  given,  the  answer  would  be  greater  than  the  third 
term ;  consequently  the  greater  of  the  two  similar  numbers  must 
then  be  taken  for  the  second  term,  and  the  less  for  the  first. 

4.  The  reason  for  multiplying  the  second  and  third  terms  together 
and  dividing  the  product  by  the  first,  is  because  the  product  of  tht 
means  divided  by  one  of  the  extremes,  gives  the  other  extreme  of 
ansicer.    (Arts.  93,  492.) 

2.  If  9  yards  of  cloth  cost  $54,  what  will  23  yards  cost  ? 
Statement.— 9  yd. :  23  yd. : :  $54 :  Ans. 

And  ($54  x  23)-5-9=$i38,  the  Ans. 

By  Cancellation. — Since  9  yds.  cost  OPERATION. 

$54,  i  yd.  will  cost  i  of  $54,  or  $\A;  $^  X  2 3  =  Ans. 

and  23  yds.  will  cost  $y  *  23=-^-^      6,  54  x  23  _ 

9  -r —  $138,  A.THS. 

=  $138,  Ans. 
Proof.— 9  yd. :  23  yd. : :  $54 :  $138 ;  for  9  x  138=23  x  54. 

3.  If  7  barrels  of  flour  cost  $56,  what  will  20  barrels  cost? 

4.  What  cost  75  bushels  of  wheat,  if  15  bushels  cost  $33? 

5.  "What  cost  150  sheep,  if  17  sheep  cost  $51  ? 

6.  If  5  Ib.  8  oz.  of  honey  cost  $1.65,  Avhat  will  20  Ib.  cost  ? 

7.  Paid  £i,  153.  6d.  for  6  pounds  of  tea:  what  must  be 
paid  for  a  chest  containing  65  Ib.  8  oz.  ? 

495.  Why  take  the  number  which  is  of  the  same  kind  as  the  answer  for 
the  third  term,  and  the  other  two  for  the  first  and  second?  When  place  tin- 
larger  of  the  other  two  numbers  for  the  second?  Why?  When  the  smaller? 
Why?  Why  does  the  product  of  the  second  and  third  terms  divided  by  the  first 
pive  the  answer?  What  is  the  arrangement  of  the  terms  in  the  form  of  a  propor- 
tion called  ?  If  the  first  and  second  terms  contain  different  denominations,  how 
proceed  ?  If  the  third  is  a  compound  number,  how  ? 


SIMPLE     PROPORTION.  313 

SIMPLE    PROPORTION    BY    ANALYSIS. 

496.  The  chief  difficulty  experienced  by  the   pupil  in   Simple 
Proportion,  lies  in  "  stating  the  question."     This  difficulty  arises 
from  a  wxnt  of  familiarity  with  the  relation  of  numbers.    He  will  be 
assisted  by  analyzing  the  examples  before  attempting  to  state  them. 

8.  If  7  bats  cost  $42,  how  much  will  12  hats  cost? 
ANALYSIS. — i  hat  is  i  S3vcnth  of  7  hats ;  therefore  i  hat  will 

cost  i  seventh  as  much  as  7  hats  ;  and  \-  of  $42  is  $6.  Again,  12  hats 
will  cost  12  times  as  much  as  i  hat,  and  12  times  $6  are  $72.  There- 
fore, 12  hats  will  cost  §72. 

Or,  7  hats  are  •?-,  of  12  hats;  therefore  the  cost  of  7  hats  is  -,T2  the 
cost  of  12  hats.  But  7  hats  cost  $42  ;  hence  $42  are  -^  the  cost  of 
12  hats.  Now,  if  -^  of  a  number  arc  $42,  |\,  of  that  number  is  J7-  of 
842,  which  is  $6 ;  and  jf  are  12  times  §6,  or  §72. 

By  Proportion. — 7  h. :  12  h. : :  $42  :  Ans.    That  is, 
7  h.  are  the  same  part  of  12  h.  as  $42  arc  of  the  cost  of  12  hats. 

497.  Solve  the  following  examples  both  by  analysis  avid 
proportion. 

9.  If  ii  men  can  cradle  33  acres  of  grain  in  i  day,  how 
many  acres  can  45  men  cradle  in  the  same  time  ? 

10.  When  mackerel  are  $150  for  12  barrels,  what  must 
I  pay  for  75  barrels  ? 

11.  How  far  will  a  railroad  car  go  in  12  hours,  if  ii 
goes  at  the  rate  of  15  miles  in  40  minutes  ? 

12.  A  bankrupt  owes  $3500,  his  assets  are  $1800:  how 
much  will  a  creditor  receive  whose  claim  is  $560  ? 

13.  At  the  rate  of  18  barrels  for  $63,  what  will  235  bar- 
rels of  apples  cost  ? 

14.  If  $250  earns  817^  interest  in  i  year,  how  much 
will  $1900  earn  in  the  same  time? 

15.  If  a  car  wheel  turns  round  6  times  in  33  yards,  how 
many  times  will  it  turn  round  in  going  7  miles  ? 

16.  If  a  clerk  can  lav  up  81500  in  i-£  year,  how  long  will 
It  take  him  to  lay  up  $5000  ? 

17.  If  6  men  can  hoe  a  field  of  corn  m  20  hours,  ho\? 
long  will  it  take  15  men  to  hoe  it? 

14 


314  SIMPLE     P  11  OP  OUT  I  ON. 

1 8.  An  engineer  found  it  would  take  75  men  220  days 
to  build  a  fort;  the  general  commanding  required  it  to  he 
built  in   15  days:   how  many  men   must   the   engineer 
employ  to  complete  it  in  the  required  time  ? 

19.  If  5  oz.  of  silk  can  be  spun  into  a  thread  100  rods 
long,  what  weight  of  silk  is  required  to  spin  a  thread  that 
will  reach  the  moon,  240000  miles  distant  ? 

20.  How  many  horses  will  it  take  to  consume  a  scaffold 
of  hay  in  40  days,  if  1 2  horses  can  consume  it  in  90  days  ? 

21.  If  |  acre  of  land  costs  $15,  what  will  25^  acres  cost  ? 

22.  If  f  of  a  ton  of  iron  costs  £f ,  what  will  ^-  of  a  ton  cost  ? 

23.  If  io£  Ib.  sugar  cost  $if,  what  will  30^  Ib.  cost  ? 

24.  If  |  of  a  chest  of  tea  costs  $35.50,  what  will  15^ 
chests  cost  ? 

25.  What  will  48!  tons  of  hay  cost,  if  1 2^  tons  cost  $126^  ? 

26.  If  ^  of  a  ship  is  worth  $16250!,  what  is  T3g-  of  it  worth  ? 

27.  What  will  it  cost  me  for  a  saddle  horse  to  go  100 
miles  if  I  pay  at  the  rate  of  37-^  cts.  for  3  miles  ? 

28.  What  must  be  the  length  of  a  slate  that  is  10  in. 
wide,  to  contain  a  square  foot  ? 

29.  How  many  yards  of  carpeting  f  yard  wide  will  it 
take  to  cover  a  floor  15  ft.  long  and  12  feet  wide  ? 

30.  If  a  man's  pulse  beats  68  times  a  minute,  how  mam 
times  will  it  beat  in  24  hours  ? 

31.  If  Halley's  comet  moves  2°  45'  in  u  hours,  how  far 
will  it  move  in  30  days  ? 

32.  If  a  pole  10  ft.  high  cast  a  shadow  i\  ft.  long,  how 
high  is  a  flag-staff  whose  shadow  is  60  ft.  long  ? 
""JijTAt  the  rate  of  3  oranges  for  7  apples,  how  many 
oranges  can  be  bought  with  150  apples? 

34.  If  an  ocean  steamer  runs  1250  miles  in  3  days  8  h., 
how  far  will  she  run  in  8  days? 

35.  If  12  men  can  harvest  a  field  of  wheat  in  n  days, 
how  many  men  are  required  to  harvest  it  in  4  days  ? 

36.  The  length  of  a  croquet-ground  is  45  feet ;  and  its 
width  is  to  its  length  as  2  to  3  :  what  is  its  width  ? 


SiilPLE     PROPORTION.  ?15 

37.  A  man's  annual  income  from  U.  S.  6s  is  $1350  when 
gold  is  1 1 2  \ :  what  was  it  when  gold  was  1 60  ? 

38.  George  has  10  minutes  start  in  a  foot-race;  and 
runs  20  rods  a  minute :  how  long  will  it  take  Henry,  who 
runs  28  rods  a  minute,  to  overtake  him? 

39.  If  the  driving  wheel  of  a  locomotive   makes  227 
revolutions  in  going  206  rods  6  ft.,  how  many  revolutions 
will  it  make  in  running  18  miles  240  rods? 

40.  If  3  Ibs.  of  coffee  cost  $1.20,  and  10  Ibs.  of  coffee  are 
worth  6  Ibs.  of  tea,,  what  will  60  Ibs.  of  tea  cost  ? 

— 3TT"37can  chop  a  cord  of  wood  in  4  hours,  and  B  in  6 
hours :  how  long  will  it  take  both  to  chop  a  cord  ? 

42.  A  reservoir  has  3  hydrants;  the  first  will  empty  it 
in  8  hours,  the  second  in  10;  the  third  in  12  hours:  if  all 
run  together,  how  long  will  it  take  to  empty  it  ? 

43.  A  man  and  wife  drank  a  keg  of  ale  in  18  d. ;  it  would 
last  the  man  30  d. :  how  long  would  it  last  the  woman  ? 

44.  A  fox  is  100  rods  before  a  hound,  but  the  hound 
runs  20  rods  while  the  fox  runs  18  rods:  how  far  must 
the  hound  run  before  he  catches  the  fox  ? 

45.  A  cistern  holding  3600  gallons  has  a  supply  and  a 
discharge  pipe ;  the  former  runs  45  gallons  an  hour,  the 
latter  33  gallons :  how  long  will  it  take  to  fill  the  cistern, 
when  both  are  running? 

46.  A  clerk  who  engaged  to  work  for  8500  a  year,  com- 
menced at  12  o'clock  Jan.  ist,  1869,  and  left  at  noon,  the 
2ist  of  May  following:  how  much  ought  he  to  receive? 

47.  A  church  clock  is  set  at  12  o'clk.  Saturday  night; 
Tuesday  noon  it  had  gained  3  min. :  what  will  be  the  true 
time,  when  it  strikes  9  the  following  Sunday  morning? 

48.  A  market-woman  bought  100  eggs  at  2  for  a  cent, 
and  another  100  at  3  for  a  cent;  if  she  sells  them  at  the 
rate  of  5  for  2  cents,  what  will  she  make  or  lose  ? 

49.  Two  persons  being  336  miles  apart,  start  at  the  same 
time,  and  meet  in  6  days,  one  traveling  6  miles  a  day 
faster  than  the  other:  how  far  did  each  travel? 


COMPOUND    PROPORTION. 


!2   ) 

>  : :  1 2  :  2,  is  a  compound  proportion. 

*  O    J 


498.  Compound  Proportion  is  an  equality  of 
a  jompound  and  a  simple  ratio.  Thus, 

4 :  2 

9 
It  is  read,  "  The  ratio  of  4  into  9  is  to  2  into  3  as  12  to  2." 

i.  If  4  men  saw  20  cords  of  wood  in  5  days,  how  many 
cords  can  1 2  men  saw  in  3  days  ? 

ANALYSIS. — The  answer  is   to  STATEMENT. 

be  in  cords ;    we  therefore  make     4  m. :  1 2  m.  )      4  , 

20  c.  the  third  term.     The  other     5  d.  :  3  d.      J    ' 
given  numbers  occur  in  pairs,  two      20  C.  X  12  X  3  =:  720  C. 
of  a  kind ;  as,  4  men  and  12  men,  4x5  =  20 

5  days  and  3  days.     We  arrange          72OC.-J-2O  —  36  C.  Ans. 
theso  pairs  in  ratios,  as  we  should 

in  simple  proportion,  if  the  answer  depended  on  each  pair  alone. 
That  is,  since  12  ra.  will  saw  more  than  4  men,  we  take  ~2  for  the 
second  term  and  4  for  the  first.  Again,  since  12  men  will  saw  lens 
in  3  days  than  in  5  days,  we  take  3  for  the  second  term  and  5  for  the 
first.  Finally,  dividing  the  product  of  the  second  and  third  terms 
20  c.  x  12  x  3=720  c.  by  the  product  of  the  first  terms  4  x  5=20,  wo 
have  36  cords  for  the  answer.  Hence,  the 

EULE. — I.  Make  that  number  tho  third  term  which  is  of 
the  same  kind  as  tho  answer. 

II.  Taka  the  other  numbers  in  pairs  of  the  same  kind, 
and  arrange  them  as  if  the  answer  depended  on  each  couplet, 
as  in  simple  proportion.     (Art.  494.) 

III.  Multiply  the  second  and  third  terms  together,  and 
divide  the  product  by  the  product  of  the  first  terms,  cancelling 
the  factors  common  to  the  first  and  second,  or  to  the  first 
and  third  terms.     Tlie  quotient  will  be  the  answer. 

PROOF. — If  the  product  of  the  first  and  fourth  terms 
equals  that  of  the  second  and  third  terms,  the  work  is  ri'jht. 

498.  What  is  Compound  Proportion  1  Explain  the  first  example.  The  rule. 
Proof.  Note.  How  proceed  when  the  first  and  second  terms  contain  different 
denominations  ?  When  the  third  does  ?  How  else  may  questions  ill  Compound 
Proportion  he  solved  1 


COMPOUND     PROPORTION. 


317 


NOTES. — i.  The  terms  of  each  couplet  in  the  compound  ratio  mu,.^ 
be  reduced  to  the  same  denomination,  and  the  third  term  to  the  lowest 
denomination  contained  in  it,  as  in  Simple  Proportion. 

2.  In  Compound  Proportion,  all  the  terms  are  given  in  couplets  or 
pairs  of  the  srtme  kind,  except  one.     This  is  called  the  odd  term,  or 
demand,  and  is  always  the  same  kind  as  the  answer. 

3.  It  should  be  observed  that  it  is  not  the  ratio  of  4  to  2,  nor  of  9  to 
3  alone  that  equals  the  ratio  of  12  to  2 ;  for,  4-4-2  =  2  and  9-7-3=3, 
while  12-5-2=6.     But  it  is  the  ratio  compounded  of  4  x  9  to  2x3, 
which  equals  the  ratio  of  12  to  2.     Thus,  (4  x  g)-^-(2  x  3)=6  ;  and 
i2-=-2=6.     (Art.  498.) 

4.  Compound  Proportion  was  formerly  called  "  Double  Rule  of 
Three." 

499.  Problems  in  Compound  Proportion  may  also  be  solved  by 
Analysis  and  Simple  Proportion.  Take  the  preceding  example : 

By  Analysis. — If  4  men  saw  20  cords  in  5  d.,  i  m.  will  saw  ^  of 
20  c.,  which  is  5  c.,  and  12  m.  will  saw  12  times  5  c.,  or  60  cords,  in 
the  same  time.  Again,  if  12  men  saw  60  c.  in  5  days,  in  i  d.  they 
will  saw  ^  of  60  c  ,  or  12  cords,  and  in  3  d.  3  times  12  c.,  or  36  cords, 
the  answer  required. 

By  Simple  Proportion. — 4  m. :  12  m. : :  20  c. :  the  cords  12  m.  will 
saw  in  5  d. ;  and  20  c.  x  12-1-4=60  c.  in  5  days.  Again,  5  d. :  3  d.  : : 
60  c. :  the  cords  12  men  will  saw  in  3  d.  And  60  c.  x  3-7-5=36  cords, 
the  same  as  before. 

2.  If  4  men  earn  §219  in  30  days,  working  10  hours  a 
day,  how  much  can  9  men  earn  in  40  clays,  working 
8  hours  a  day? 

By  Cancellation. 

9m.  3 

5, 15,  S&  d.  ' 
IBh. 


STATEMENT. 

4m.:  9  m. 
30  d.  :  40  d. 
10  h.  :  8  h. 

r  :  :  $219  :  Ans. 

(§219x9x4 

OX  8)  -T- 

=  $525£,  Ans. 


5 


Bh.  4 


x  4  x  3  = 


3.  If  6  men  can  mow  28  acres  in  2  days,  how  long  will 
it  take  7  men  to  mow  42  acres  ? 

4.  If  8  horses  can  plow  32  acres  in'  6  days,  how  many 
hcrses  will  it  take  to  plow  24  acres  in  4  days  ? 

5.  If  the  board  of  a  family  of  8  persons  amounts  to  $300 
in  15  weeks,  how  long  will  81000  board  12  persons? 


318  COMPOUND     PHOPOBTIOX. 

6.  "What  will  be  the  cost  of  28  boxes  of  candles  con  tain - 
'ng  20  pounds  apiece,  if  7  boxes  containing  15  pounds 
ipiece  can  be  bought  for  $23.75  ? 

7.  If  the  interest  of  $300  for  10  months  is  $20,  what 
\vill  be  the  interest  of  $1000  for  15  months? 

8.  If  a  man  walks  180  miles  in  6  days,  at  10  h.  each, 
how  many  miles  can  he  walk  in  15  days,  at  8  h.  each  ? 

9.  If  it  costs  $160  to  pave  a  sidewalk  4  ft.  wide  and 
40  ft,  long,  what  will  it  cost  to  pave  one  6  ft.  wide  and 
125  ft.  long? 

10.  If  it  requires  800  yards  of  cloth  £  yd.  wide  to  supply 
100  men,  how  many  yards  that  is  -g-  wide  will  it  require  to 
clothe  1500  men? 

n.  If  75  men  can  build  a  wall  50  ft.  long,  8  ft.  high,  and 
3  ft.  thick,  in  10  days,  how  long  will  it  take  100  men  to 
build  a  wall  150  ft.  long,  10  ft.  high,  and  4  ft.  thick  ? 

12.  If  it  costs  $56  to  transport  7  tons  of  goods  1 10  miles, 
how  much  will  it  cost  to  transport  40  tons  500  miles  ? 

1 3.  If  30  Ib.  of  cotton  will  make  3  pieces  of  muslin  42  yds. 
long  and  f  yd.  wide,  how  many  pounds  will  it  take  to 
make  50  pieces,  each  containing  35  yards  i£  yd.  wide  ? 

14.  If  the  interest  of  $600,  at  7^,  is  $35  for  10  months, 
what  will  be  the  int.  of  $2500,  at  6r/c,  for  5  months? 

15.  If  9  men,  working  10  hours  per  day,  can  make  18 
sofas  in  30  days,  how  many  sofas  can  50  men  make  in 
90  days,  working  8  hours  per  day  ? 

1 6.  If  it  takes  9000  bricks  8  in.  long  and  4  in.  wide  to 
pave  a  court-yard  50  ft.  long  by  40  ft.  wide,  how  many 
tiles  10  in.  square  will  be  required  to  lay  a  hall-floor  75  ft. 
long  by  8  ft.  wide  ? 

17.  If  in  8  days  15  sugar  maples,  each  running  12  quarts 
of  sap  per  day,  make  10  boxes  of  sugar,  each  weighing 
6  Ib.,  how  many  boxes  weighing  10  Ib.  apiece,  will  a  maple 
grove  containing  300  trees,  make  in  36  days,  each  tree 
running  16  quarts  per  day?  Ans.  720  boxes. 


PAJlTiTlVil     PKOPOIITIOX.  319 

PARTITIVE    PROPORTION. 

500.  Partitive  Proportion  is  dividing  a  number 
into  two  or  more  parts  having  a  given  ratio  to  each  other. 

501.    To    divide   a   Number   into   two    OP   more   parts  which 
shall  be  proportional  to  given  numbers. 

1.  A  and  B  found  a  purse  of  money  containing  $35, 
which  they  agree  to  divide  between  them  in  the  ratio  of  2 
to  3 :  how  many  dollars  will  each  have  ? 

By  Proportion.— The   sum  of  the  STATEMENT. 

proportional  parts  is  to  each  separate  5  :  2  :  :  $35  :  A's  S. 

part  as  the  number  to  bo  divided  is  to  r  .  ,  .  .  *- -  :  J$'g  g€ 

each  man's  share.     That  is,  5  (2  +  3) 

is  to  2  as  $35  to  A's  share.     Again,  5      ('$35  X  2 )  -^  5  =!f> 1 4  A  S  S. 
is  to  3  as  $35  to  B's  share.    Hence,  the     ($35  X  3)  -r-  5  =•  $2 1  B's  S. 

EULE. — I.  Take  the  number  to  be  divided  for  the  third 
term ;  each  proportional  part  successively  for  the  second 
1 1'  /•//?;  and  their  sum  for  the  first. 

II.  The  product  of  the  second  and  third  terms  of  each 
proportion,  divided  by  the  first,  will  be  the  corresponding 
part  required. 

By  Analysis. — Since  A  had  2  parts  and  B  3,  both  had  2  +  3,  or  5 
parts.  Hence,  A  will  have  I  and  B  f  of  the  money.  Now  £  of  $35 
—$14,  and  |  of  $35  =  $21.  Hence,  the 

RULE. — Divide  the  given  number  by  the  sum  of  the  pro- 
portional numbers,  and  multiply  the  quotient  by  each  one's 
proportional  part. 

2.  It  is  required  to  divide  78  into  three  parts  which 
shall  be  to  each  other  as  3,  4,  and  6.     Ans.  18,  24,  36. 

3.  A  man  having  200  sheep,  wished   to   divide  them 
into  three  flocks  which  should  be  to  each  other  as  2,  3, 
and  5  :  how  many  will  each  flock  contain  ? 

4.  A  miller  had  250  bushels  of  provender  composed  of 
oats,  peas,  and  corn  in  the  proportion  of  3,  4,  and  5^-:  how 
many  bushels  Avere  there  of  each  kind  ? 

tan.  How  divide  a  number  into  parts  having  u  irlvtm  r»ti>.  > 


320  PARTNERSHIP. 

5.  A  father  divided  497  acres  of  land  among  his  four 
sons  in  proportion  to  their  ages,  which  were  as  2,  3,  4,  and 
5  :  how  many  acres  did  each  receive  ? 

502.  The  principles  of  Partitive  Proportion  are  appli- 
cable to  those  classes  of  problems  commonly  arranged 
under  the  heads  of  Partnership,  Bankruptcy,   General 
Average,  etc.,  in  which  a  given  number  is  to  be  divided 
into  parts  having  a  given  ratio  to  each  other. 

."?• ' 

PAETj^EESHIP. 

503.  Partnership  is  the  association  of  two  or  more 
persons  in  business  for  their  common  profit. 

It  is  of  two  kinds ;  Simple  and  Compound. 

504.  Simple  Partnership  is  that  in  which  the 
capital  of  each  partner  is  employed  for  the  same  time. 

505.  Compound  Partner  sh  fp  is  that  in  which 
the  capital  is  employed  for  unequal  times. 

NOTE. — The  association  is  called  a  firm,  Jiousc,  or  company ;  and 
the  persons  associated  arc  termed  partners. 

506.  The  Capital  is  the  money  or  property  employed 
in  the  business. 

The  Profits  arc  the  gains  shared  among  the  partners, 
and  are  called  dividends. 

NOTES. — i.  The  profits  are  divided  as  the  partners  may  agroo. 
Other  things  being  equal,  when  the  capital  is  employed  for  the  .<."///,> 
time,  it  is  customary  to  divide  the  profits  according  to  the  amount 
of  capital  each  one  furnishes. 

2.  When  the  capital  is  employed  for  unequal  times,  the  profits  are 
usually  divided  according  to  the  amount  of  capital  each  furn i 
and  the  time  it  is  employed. 

502.  To  what  arc  Ihe  principles  of  Partitive  Proportion  applicable  ?  503.  \Vliat 
Is  Partnership?  Of  how  many  kinds?  504.  Simple  Partnership?  505.  Com- 
pound? Note.  What  is  the  association  called?  506.  What  is  the  capital  ?  Th« 
profits  ?  What  called  f 


PARTNERSHIP.  32i 

PRORLEM    I. 

507.  To  find  each  Partner's  Share  of  the  Profit  or  Loss, 
when  divided  according  to  their  capital. 

1.  A  and  B  entered  into  partnership;  the  former  furnish- 
ing 8648,  the  latter  $1080,  and  agreed  to  divide  the  profit 
according  to  their  capital.     They  made  $432 :  what  was 
each  one's  share  of  the  profit  ? 

ANALYSIS. — The  capital  equals  $648  +  $io8o=$i728. 

$1728  (capital) :    $648  (A's  cap.) : :  $432  (profit) :  A's  share,  or  $162. 

$1728        "        :  $1080  (B's  cap  )::  $432      "       :  B's  share,  or  $270. 

Or  thus:  The  profit  $432-r-$i728  (the  cap.) =.2 5 ;  that  is,  the 
profit  is  25  fe  of  the  capital.  Therefore  each  man's  share  of  the 
profit  is  25$  of  his  capital.  Now  $648  x  .25  =  $i62,  A's  share;  and 
$1080  x  .25=8270,  B's  share.  Hence,  the 

RULE. — The  whole  capital  is  to  each  partner's  capital,  as 
the  whole  profit  or  loss  to  each  partner's  share  of  the  profit 
or  loss.  • 

Or,  Find  what  per  cent  the  profit  or  loss  is  of  the  whole 
capital,  and  multiply  each  man's  capital  by  it.  (Art.  339.) 

2.  A  and  B  form  a  partnership,  A  furnishing  $1200, 
and  B  81500;  they  lose  8500:  what  is  each  one's  share  of 
the  loss  ? 

3.  A  and  B  buy  a  saw-mill,  A  advancing  $3000,  and 
B  $4500;  they  rent  it  for  $850  a  year:  what  should  each 
receive  ? 

4.  The  net  gains  of  A,  B,  and  C  for  a  year  are  $12500 ; 
A  furnishes  $15000,  B  Si2'ooo,  and  C  $10000:  how  should 
the  p*rofit  be  divided  ? 

5.  Three   persons  entering  into  a  speculation,  made 
$15300,  which  they  divided  in  the  ratio  of  2,  3,  and  4: 
how  much  did  each  receive? 

6.  A,  B,  and  C  hire  a  pasture  for  $320  a  year;  A  put  in 
80,  B  1 20,  and  C  200  sheep :  how  much  ought  each  man 
to  pay  ? 

507.  How  find  each  partner's  profit  or  loss,  when  their  capital  is  employed  the 
same  time  ? 


3'2  PARTNERSHIP. 

PROBLEM    II. 

508.  To  find  each   Partner's   Share  of  the  Profit  or  Loss, 
when  divided  according  to  capital  and  time. 

7.  A  and  B  enter  into  partnership ;  A  furnishes  $400 
for  8  months,  and  B  $600  for  4  months;  they  gain  $350: 
what  is  each  one's  share  of  the  profit  ? 

ANALYSIS. — In  this  case  the  profit  of  the  partners  depends  on  two 
conditions,  viz. :  the  amount  of  capital  each  furnishes,  and  the  time 
it  is  employed. 

But  the  use  of  $400  for  8  months  equals  that  of  8  times  $400,  or 
$3200,  for  I  m. ;  and  $600  for  4  in.  equals  4  times  $600,  or  $2400,  for 
i  m.  The  respective  capitals,  then,  are  equivalent  to  $2400  and 
$3200,  each  employed  for  I  m.  Now,  as  A  furnished  $3200,  and 
B  $2400,  the  whole  capital  equals  $3200  +  $2400=35600.  Therefore, 

$5600  :  $3200  : :  $350  (profit) :  A's  share,  or  $200. 
$5600  :  $2400 : :  $350      "       :  B's  share,  or  $150. 

Or  thus:  The  gain  $350  -4-  $5600  (the  cap.)  =  .06^,  or  f>\%. 
Therefore,  $3200 x  .06}  =  $200,  A's  share;  and  $2400 x  .06;  =  $150, 
B's  share.  Hence,  the 

RULE. — Multiply  each  partner's  capital  by  the  time  it  is 
employed.  Consider  these  products  as  their  respective  capi- 
tals, and  proceed  as  in  the  last  problem. 

Or,  Find  what  per  cent  the  profit  or  loss  is  of  the  whole 
capital,  and  multiply  each  man's  capital  by  it.  (Art,  339.) 

NOTE. — The  object  of  multiplying  each  partner's  capital  by  the 
time  it  is  employed  is,  to  reduce  their  respective  capitals  to  equivalents 
for  the  same  time. 

8.  A,  B,  and  C  form  a  partnership;  A  furnishing  $500 
for  9  m.,  B  $700  for  i  year,  and  C  $400  for  15  months; 
they  lose  8600 :  what  is  each  man's  share  of  the  loss  ? 

9.  Two  men  hire  a  pasture  for  $50 ;  one  put  in  20  horses 
for  12  weeks,  the  other  25  horses  for  10  weeks:  how  much 
shoiild  each  pay  ? 

508.  How  find  each  one's  share,  when  their  capital  is  employed  for  unequal 
time*  ?  .Y<?te ,  Why  multiply  each  one's  capital  by  the  time  it  is  employed  ? 


BANKRUPTCY.  323 

to  A.  B,  C,  and  D  commenced  business  Jan.  ist,  1870,- 
when  A  .furnished  $iooo;/  March  ist,  B  put  in  $1200; 
July  ist,  C  put  in  $1500;  and  Sept.  ist,  D  put  in  $2000; 
during  the  year  they  made  $1450 :  how  much  should  each 
receive  ? 

n.  A  store  insured  at  the;  Howard  Insurance  Co.  for 
$3000;  in  the  Continental,  $4500;  in  the  American, 
&6ooo,  was  damaged  by  fire  to  the  amount  of  $6750: 
what  share  of  the  loss  should  each  company  pay  ? 

REMARK. — The  loss  should  be  averaged  in  proportion  to  the  risk 
assumed  by  each  company. 

12.  A  quartermaster  paid  $2500  for  the  transportation 
of  provisions ;  A  carried  150  barrels  40  miles,  B  1 70  barrels 
60  miles,  C  210  barrels  75  miles,  and  D  250  barrels  100 
miles :  how  much  did  he  pay  each  ? 

13.  A,  B,  and  C,  formed  a  partnership,  and  cleared\ 
$12000;  A  put  in  $8000  for  4  m.,  and  then  added  $2000 
for  6  m. ;  B  put  in  $16000  for  3  m.,  and  then  withdrawing 
half  his  capital,  continued  the  remainder  5  m.  longer; 
C  put  in  $13500  for  7  m. :  how  divide  the  profit. 


BANKRUPTCY. 

509.  Bankruptcy  is  inability  to  pay  indebtedness. 
NOTE. — A  person  unable  to  pay  his  debts  is  said  to  be  insolvent, 

and  is  called  a  bankrupt. 

510.  The  Assets  of  a  bankrupt  are  the  property  in 
his  possession. 

The  Liabilities  are  his  debts. 

511.  The  Net  Proceeds  are  the  assets  less  the  ex- 
pense of  settlement.    They  are  divided  among  the  creditors 
according  to  their  claims. 

509.  What  ia bankruptcy?  510.  What  arc  assets?  Liabilities?  511.  Net  pro- 
ceeds? 512.  How  find  each  creditor's  dividend,  wbeu  the  liabilities  a«d  the  u«i 
proceeds  are  given? 


324  ALLIGATION. 

o 

512.    To   find   each   Creditor's    Dividend,   the    Liabilities   and 
Net  Proceeds  being  given. 

i.  A  merchant  failed,  owing  B  81260,  0  $1800,  and 
D  81940;  his  assets  were  $1735,  and  the  expenses  of 
settling  $435  :  how  much  did  each  creditor  receive  ? 

ANALYSIS. — The  liabilities  are  $1260  +  $1800  +  $1940=15000;  and 
the  uet  proceeds  $1735  —  $435=81300.  Now, 

$5000  :  $1260  : :  $1300  :  B's  divideud,  or  $327.60. 
$5000  :  $1800:  .$1300:  C's         "         or  $468.00. 
$5000 :  $1940 : : 81300 :  D's         "         or  $504  40. 
Or  thus:  The  net  proceeds  $1300 -=-$5000=  26,  or  26%,  the  rate 
he  is  able  to   yay.     (Art.  339.)    Now   $1260  x  .26  =  $327.60  B's; 
§1800  x  .26=0468  C's ;  $1940  x  .26=§50440  D's.     Hence,  the 

RULE, — T/ic  wliolo  liabilities  arc  to  each  creditor's  claim, 
as  the  net  proceeds  to  each  creditor's  dividend. 

Or,  Find  u'hat  per  cent  the  net  proceeds  are  of  the 
liabilities,  and  multiply  each  creditor's  claim  by  it. 

2.  A  bankrupt  owes  A  $6300,  B  84500,  and  D  83200; 
his  assets  are  $5250,  and  the  expenses  of  settling  §1500: 
how  much  will  cacli  creditor  receive  ? 

3.  A.  B.  &  Co.  Avcnt  into  bankruptcy,  owing  §48400, 
and  having  $13200  assets;   the  expense   of  settling  was 
$i  100.     What  did  D  receive  on  $8240  ? 


ALLIGATION. 

513.  Alligation  is  of  two  kinds,  Medial  vx\&  Alternate. 

All  if/at  ion  Medial  is  the  method  of  finding  the 
mean  value  of  mixtures. 

Allif/ation  Alternate  is  the  method  of  finding 
the  proportional  parts  of  mixtures  having-  a  given  value. 

NOTES. — i.  Ailigntion,  from  the  Latin  alligo,  to  tie,  or  bind 
together,  is  so  called  from  the  manner  of  connecting  the  ingredients 
by  curve  lin^s  in  some  of  the  operations. 

2.  Alternate,  Latin  nlternntus,  by  turns,  refers  to  the  manner  of 
connecting  the  prices  above  the  moan  pries  with  those  bdow. 

3.  The  term  mediftl  is  from  the  Latin  median,  middle  or  average. 

513.  What  i?  alligation  medial  ?    Alternate-  't 


ALLIGATION.  32£ 

ALLIGATION     MEDIAL. 

514.  To  find  the  Mean  Value  of  a  mixture,  the  Price  and 
Quantity  of  each  ingredient  being  given. 

1.  Mixed  50  Ibs.  of  tea  at  90  cts.,  60  Ibs.  at  $1.10,  and 
80  Ibs.  at  $1.25  :  what  is  the  mean  value  of  the  mixture  ? 

ANALYSIS— The  total  value  of  the  first  $.90  X  50=    $45.00 

kind  .90  x  so=$45.oo,  the  second  $i.iox  60  $1. 10x60=    $66.00 

=  $6600,  the   third    $1.25  x  80= $100.00  ;  $1.25  x8o=$ioo.oo 

therefore  the  total  value  of  the  rnixture=:  190  )  $21  i.oo 

$211.00.    But  the  quantity  raixed^igolbs.  4<no~& —       , 
Now,  if  190  Ibs.  are  worth  $211,  i  Ib.  is 

worth  T£TT  of  $211,  or  $i.n  +  .    Therefore  the  mean  value  of  the 
mixture  is  $1.11 4- per  pound.     Hence,  the 

KULE. — Divide  the  value  of  the  whole  mixture  ~by  the  sum 
of  the  ingredients  mixed. 

NOTE. — If  an  ingredient  costs  nothing,  as  water,  sand,  etc.,  its 
value  is  o ;  but  the  quantity  itself  must  ba  added  to  the  other 
ingredients. 

2.  If  I  mix  3  kinds  of  sugar  worth  12,  15,  and  20  cts.  a 
pound,  what  is  a  pound  of  the  mixture  worth  ? 

3.  A  farmer  mixed  30  bu.  of  corn,  at  $1.25,  with  25  bu. 
of  oats,  at  60  cts.,  and  10  bu.  of  peas,  at  95  cts. :  what  was 
the  average  value  of  the  mixture  ? 

4.  A  grocer  mixed  5  gal.  molasses,  worth  80  cts.  a  gal., 
and  107  gallons  of  water,  with  a  hogshead  of  cider,  at  20 
cts. :  what  was  the  average  worth  of  the  mixture  ? 

5.  A  goldsmith  mixed  12  oz.  of  gold  22  carats  fine  with 
8  oz.  20  carats,  and  7  oz.   18  carats  fine:  what  was  the 
average  fineness  of  the  composition  ? 

6.  A  milkman  bought  40  gallons  of  new  milk,  at  4  cts. 
a  quart,  and  60  gallons  of  skiriimed  milk  at  2  cts.  a  quart, 
which  he  mixed  with  1 2  gallons  of  Avater,  and  sold  tho 
whole  at  6  cts.  a  quart :  required  his  profit  ? 

Note.  Why  is  this  rule  called  alligation  ?  Why  alternate  ?  Wh.it  is  the  import 
of  medial?  514.  How  find  the  mean  value  of  a  mixture,  when  the  price  and 
quantity  are  giver.  ? 


326  ALLIGATION     ALTEENATE. 

ALLIGATION     ALTERNATE. 
PROBLEM    I. 

515.    To  find  the  Proportional  Parts  of  a   Mixture,  the 
Mean  Price  and  the  Price  of  each  ingredient  being  given. 

7.  A  grocer  desired  to  mix  4  kinds  of  tea  worth  33.,  8s., 
i  is.,  and  1 28.  a  pound,  so  that  the  mixture  should  be  worth 
93.  a  pound :  in  what  proportion  must  they  be  taken  ? 

ANALYSIS. — To  equalize  the  gain  and  loss,  we  OPERATION. 

compare  the  prices  in  pairs,  one  being  above  and          3 

the  other  below  the  mean  price,  and,  for  conve- 
nience, connect  them  by  curve  lines.    Taking  the 


first  and  fourth ;  on  i  Ib.  at  33.  the  gain  is  6s. ;        j  2 ^   6 

on  i  Ib.  at  I2s.  the  loss  is  33.,  which  we  place  op- 
posite the  12  and  3.  Therefore,  it  takes  i  Ib.  at  33.  to  balance  the 
loss  on  2  Ib.  at  123.,  and  the  proportional  parts  of  this  couplet  are 
as  i  to  2,  or  as  3  to  6.  But  3  and  6  are  the  differences  between  the 
mean  price  and  that  of  the  teas  compared,  taken  inversely. 

Again,  i  Ib.  at  8s.  gains  is.,  and  i  Ib.  at  us.  loses  2s.,  which  we 
place  opposite  the  u  and  8.  Therefore,  it  takes  2  Ib.  at  8s.  to  bal- 
ance the  loss  on  i  Ib.  at  us.,  and  the  proportional  parts  of  this 
couplet  are  as  2  to  I.  But  2  and  i  are  the  differences  between  the 
mean  price  and  that  of  the  teas  compared,  taken  inversely.  The 
parts  are  3  Ibs.  at  33.,  2  Ibs.  at  8s.,  i  Ib.  at  us.,  6  Ibs.  at  123. 

If  we  compare  ihc  first  and  third,  the  second  and  fourth,  the  pro- 
portional parts  will  be  2  Ibs.  at  33.,  3  Ibs.  at  8s.,  6  Ibs.  at  us.,  and 
i  Ib.  at  128.  Hence,  the 

KULE. — I.  Write  the  prices  of  the  ingredients  in  a  column, 
with  the  mean  price  on  the  left,  and  taking  them  in  pairs, 
one  less  and  the  other  greater  than  the  mean  price,  connect 
them  by  a  curve  line. 

II.  Place  the  difference  between  the  mean  price  and  that 
of  each  ingredient  opposite  the  price  with  which  it  is  com- 
pared. The  sum  of  the  differences  standing  opposite  each 
price  is  the  proportional  part  of  that  ingredient. 

515.  How  find  the  proportional  parts  of  a  mixture,  when  the  mean  price  and. 
the  price  of  each  Ingredient  are  given  ? 


ALLIGATION     ALTERNATE.  327 

REM. — Since  the  results  show  the  proportional  parts  to  be  taken, 
it  follows  if  each  is  multiplied  or  divided  by  the  same  number,  an 
endless  variety  of  answers  may  be  obtained. 

20  METHOD. — Since  the  mean  price  is  gs.  a  pound, 

)l  Ib.  at  33.  gains  6s. ;  hence,  to  gain  is.  takes  £  lb.=£.  Ib. 
i  Ib.  at  8s.  gains  is. ;  hence,  to  gain  is.  takes  i  lb.=£  Ib. 
1  i  Ib.  at  us.  loses  2s. ;  hence,  to  lose  is.  takes  %  lb.=|  Ib. 
'    i  Ib.  at  I2S.  loses  33. ;  hence,  to  lose  is.  takes  5  Ib.rrf  Ib. 
Reducing  these  results  to  a  common  denominator,  and  using  the 
numerators,  the  proportional  parts  are  i  Ib.  at  33.,  6  Ibs.  at  8s., 
3  Ibs.  at  us.,  and  2  Ibs.  at  123.    Hence,  the 

RULE. — Take  the  given  prices  in  pairs,  one  greater,  the 
other  less  than  the  mean  price,  and  find  liow  much  of  each 
article  is  required  to  gain  or  lose  a  unit  of  the  mean  price, 
setting  the  result  on  the  right  of  the  corresponding  price. 

Reduce  the  results  to  a  common  denominator,  and  the 
numerators  will  be  the  proportional  parts  required. 

NOTES. — i.  If  there  are  three  ingredients,  compare  the  price  of 
the  one  which  is  greater  or  less  than  the  mean  price  with  each  of 
the  others,  and  take  the  sum  of  the  two  numbers  opposite  this  price. 

2.  The  reason  for  considering  the  ingredients  in  pairs,  one  above, 
and  the  other  below  the  mean  price,  is  that  the  loss  on  one  may  be 
counterbalanced  by  the  gain  on  another. 

[For  Canfield's  Method,  see  Key  to  New  Practical.] 

8.  A  miller  bought  wheat  at  $1.60,  $2.10,  and  $2.25  per 
bushel  respectively,  and  made  a  mixture  worth  $^a  bushel: 
how  much  of  each  did  he  buy? 

9.  A  refiner  wished  to  mix  4  parcels  of  gold  15,  18,  21, 
and  22  carats  fine,  so  that  the  mixture  might  be  20  carats 
fine :  what  quantity  of  each  must  he  take  ? 

10.  A  grocer  has  three  kinds  of  spices  worth  32,  40,  and 
45  cts.  a  pound :  in  what  proportion  must  they  be  mixed, 
that  the  mixture  may  be  worth  38  cts.  a  pound? 

n.  A  grocer  mixed  4  kinds  of  butter  worth  20  cts., 
27  cts.,  35  cts.,  and  40  cts.  a  pound  respectively,  and  sold 
the  mixture  at  42  cts.  a  pound,  whereby  he  made  10  cts, 
a  pound :  how  much  of  a  kind  did  he  mix  ? 


328  ALLIGATION     ALTERNATE. 


PROBLEM    II. 

516.  When  one  ingredient,  the  Price   of  each,  and  the  Mean 
Price  of  the  Mixture  are  given,  to  find  the  other  ingredients. 

12.  How  much  sugar  worth  12,  14,  and  21  cts.  a  pound 
must  be  mixed  with  12  Ibs.  at  23  cts.  that  the  mixture 
may  be  worth  18  cts.  a  pound? 

ANALYSIS. — If  neither  ingredient  were  limited, 
the  proportional  parts  would  be  3  Ibs.  at  12  cts., 
5  Ibs.  at  14  cts.,  6  Ibs.  at  21  cts.,  and  4  Ibs.  at 


12 v     3 

14 — 0  5 
21 — J\  6 


23  cts.  23  —^    4 

But  the  quantity  at  23  cts.  is  limited  to  12  Ibs., 
which  is  3  times  its  difference  4  Ibs.  Now  the  ratio  of  12  Ibs.  to 
4  Ibs.  is  3.  Multiplying  each  of  the  proportional  parts  found  by  3 
the  result  will  be  9  Ibs.,  15  Ibs.,  18  Ibs.,  and  12  Ibs.  Hence,  the 

EULE.— Find  the  proportional  parts  as  if  the  quantity 
of  neither  ingredient  were  limited.  (Art.  515.) 

Multiply  the  parts  thus  found  oy  the  ratio  of  the  given 
ingredient  to  its  proportional  part,  and  the  products  will 
be  the  corresponding  ingredients  required. 

NOTE. — When  the  quantities  of  two  or  more  ingredients  are  given, 
find  the  average  value  of  them,  and  considering  their  sum  as  one 
juantity,  proceed  as  above.  (Art.  515.) 

13.  How  much  barley  at  40  cts.,  and  corn  at  80  cts., 
must  be  mixed  with  .  IP.  bu.  of  oats  at  30  cts.  and  20  bn. 
of  rye  at  60  cts.,  that  the  mixture  may  be  55  cts.  a  bu.  ? 

SUGGESTION. — The  mean  value  of  10  bu.  of  oats  at  30  cts.  an(' 
20  bu.  of  rye  at  60,  is  50  cts.  a  bu.  Ans.  30  bu.  barley,  and  24  bu.  corn. 

14.  How  much  butter  at  40,  45,  and  50  cts.  a  pound 
respectively,  must  I  mix  with  30  Ibs.  at  65  cts.  that  the 
mixture  may  be  worth  60  cts.  a  pound? 

15.  How  many  quarts  of  milk,  worth  4  and  6  cts.  a  quart 
respectively,  must  be  mixed  with  50  quarts  of  water  s<. 
that  the  mixture  may  be  worth  5  cts.  a  quart  ? 

516.  When  one  ingredient,  the  price  of  each  and  mean  price  are  given,  hoM 
End  the  other  ingredients  ? 


ALLIGATION     ALTERNATE.  329 

PROBLEM    III. 

517.    To   find   the    Ingredients,   the    Price    of  each,   the 
Quantity  mixed,  and  the  Mean  Price  being  given. 

1 6.  How  much  water  must  be  mixed  with  two  kinds  of 
Bourbon  costing  $4  and  $6  a  gal.,  to  make  a  mixture  of 
T5o  gal.  the  mean  price  of  which  shall  be  $3  a  gal.  ? 

ANALYSIS. — The  price  of  the  water  is  o.  Disregarding  the  quantity 
to  be  mixed,  and  proceeding  as  in  Problem  I.,  the  proportional  part? 
are  4  g.  water,  3  g.  at  $4,  and  3  g.  at  $6,  the  sum  of  which  is  4  g.  , 

3  g.  +  3  g.  — 10  gallons. 

But  the  whole  mixture  is  to  be  150  gallons.    Now  the  ratio  of 
150  g.  to  10  g.  equals  Lp$ ,  or  15. 
Multiplying  each  of  the  parts  previously  obtained  by  15,  we  have 

4  gal.  x  15=60  gal.  water;  3  gal.  x  15=45  £al.  at  $4 ;  and  3  gal.  x 
15=45  ?al.  at  $6-     Hence,  the 

RULE. — Find  the  proportional  parts  without  regard  tc 
the  quantity  to  be  mixed,  as  in  Problem  /. 

Multiply  each  of  the  proportional  parts  thus  found  by 
the  ratio  of  the  given  mixture  to  the  sum  of  these  parts,  and 
the  several  products  will  be  the  corresponding  ingredients 
required. 

17.  A  grocer  mixed  100  Ib.  of  lard  worth  6,  8,  and  12  cts. 
a  pound,  the  mean  value  of  the  mixture  being  10  cts. : 
how  many  pounds  of  each  kind  did  he  take  ? 

1 8.  Having  coffees  worth  28,  30,  38,  and  42  cts.  a  pound 
respectively,  I  wish  to  mix  200  Ibs.  in  such  proportions 
that  the   mean  value  of  the   mixture   shall   be   36   cts. 
a  pound :  how  many  pounds  of  each  kind  must  I  take  ? 

19.  A  grocer  wished  to  mix  4  kinds  of  petroleum  worth 
40,  45,  50,  and  60  cts.  a  gal.  respectively :  how  much  of 
each  kind  must  he  take  to  make  a  mixture  of  300  gallons, 
worth  52  cts.  a  gallon? 

517.  How  find  the  in^rcdu-uts.  when  the  price  of  each,  the  quantity  mixed, 
nod  the  mean  price  a.-c  gi'veii? 


INVOLUTION. 

518.  Involution  is  finding  a  power  of  a  number. 

A  Poiver  is  the  product  of  a  number  multiplied  iutc 
itself.  Thus,  2x2  —  4;  3  x  3— 9>  etc.,  4  and  9  are  powers. 

519.  Powers  are  divided  into  different  degrees;  as,  first, 
second,  third,  fourth,  etc.     The  name  shows  how  many 
times  the  number  is  taken  as  a  factor  to  produce  the  power. 

520.  The  First  Poiver  is  the  root  or  number  itself. 
The  Second  Poiver  is  the  product  of  a  number 

taken  twice  as  a  factor,  and  is  called  a  square. 

The  Third  Power  is  the  product  of  a  number  taken 
three  times  as  a  factor,  and  is  called  a  cube,  etc. 

NOTES. — i.  The  second  power  is  called  a  square,  because  the  pro- 
cess of  raising  a  number  to  the  second  power  is  similar  to  that  of 
finding  the  area  of  a  square.  (Art.  243.) 

2.  The  third  power  in  like  manner  is  called  a  cube,  because  the 
process  of  raising  a  number  to  the  third  power  is  similar  to  that  of 
finding  the  contents  of  a  cube.  (Art.  249.) 

521.  Powers  are  denoted  by  a  small  figure  placed  above  the 
number  on  the  right,  called  the  index  or  exponent ;  because  it  shows 
how  many  times  the  number  is  taken  as  a  factor,  to  produce  the  power. 

NOTE. — The  term  index  (plural  indices),  Latin  indicere,  to  proclaim. 
Exponent  is  from  the  Latin  exponere,  to  represent.    Thus, 

2]  =  2,  the  first  power,  which  is  the  number  itself. 

2'2:=2  x  2,  the  second  power,  or  square. 

23=2  x  2  x  2,  the  third  power,  or  cube. 

24=2  x  2  x  2  x  2,  the  fourth  power,  etc. 

522.  The  expression  24  is  read,  "2  raised  to  the  fourth  power,  or 
the  fourth  power  of  2." 

1.  Read  the  following,  g5,  I27,  25",  245",  381'°,  4&515,  looo24. 

2.  63  x  y4,  25"  x  48',  I408— 75s,  256'° -^  97'. 


518.  What  is  involution  ?  A  power .  519.  How  are  powers  divided  ?  520.  What 
is  the  fir^t  power  ?  The  second  ?  Third  ?  Note.  Why  is  the  second  power  called 
•  square?  Why  the  third  a  cube ?  521.  How  are  powers  denot«dy 


INVOLUTION.  331 

3.  Express  the  4th  power  of  85.  5.  The  7th  power  of  340. 

4.  Express  the  5th  power  of  348.        6.  The  8th  power  of  561. 

623.  To  raise  a  Number  to  any  required  Power. 

7.  What  is  the  4th  power  of  3  ? 

ANALYSIS. — The  fourth,  power  is  the  product  of  a  number  into 
itself  taken  four  times  as  a  factor,  and  3  x  3  x  3  x  3=81,  the  answer 
required.  Hence,  the 

RULE. — Multiply  the  number  into  itself,  till  it  is  taken 
as  many  times  as  a  factor  as  there  are  units  in  the  index 
of  the  required  power. 

NOTES. — I.  In  raising  a  number  to  a  power,  it  should  be  observed 
that  the  number  of  multiplications  is  always  on,e  less  than  the  number 
of  times  it,  is  U'jen  as  a  factor;  and  therefore  one  less  than  the 
number  of  the  index.  Thus,  43=4  x  4  x  4,  the  4  is  taken  three  times 
as  a  factor,  but  there  are  only  two  multiplications. 

2.  A  decimal  fraction  is  raised  to  a  power  by  multiplying  it  into 
itself,  and  pointing  off  as  many  decimals  in  each  power  as  there  are 
decimals  in  the  factors  employed.     Thus,  .I2=.oi,  .23=.oo8,  etc. 

3.  A  common  fraction  is  raised  to  a  power  by  multiplying  each 
term  into  itsel f.     Thus,  (\)*=  ,26 . 

4.  A  mixed  number  should  be  reduced  to  an  improper  fraction,  or 
the  fractional  part  to  a  decimal;  then  proceed  as  above.     Thus,  (2i)2 
=(5)2=**;  or  21=2.5  and  (2.5)2=6.25. 

5.  All  powers  of  I  are  I ;  for  i  x  i  x  i,  etc.=i. 

Compare  the  square  of  the  following  integers  and  that 
of  their  corresponding  decimals  : 

8.    S>  6>  7>  8>  9>   I0>  20>  3°>  4°>  5°>  6o>  7°>  8o>  9°- 
9-  -5>  -6>  -7>  -s>  -9>  -OI>  -02>  -°3>  -04?  -°5>  -°6>  -°7>  -°S,  .09. 

Raise  the  following  numbers  to  the  powers  indicated : 


io.  53- 

13-  4s- 

16.  2.O33. 

19-  I4- 

II.    26. 

14.  84. 

17.  4.00033. 

20.  -£3. 

12.    I323. 

J5-  253- 

1  8.  400.05  3. 

21.    2-^4. 

5_'3.  How  raise  a  number  to  a  power?  Note.  In  raising  a  number  to  a  power, 
how  many  multiplications  are  there  ?  How  is  a  decimal  raised  to  a  power  ?  A 
common  fraction'/  A  mixed  number?  524.  How  find  the  product  of  two  or 
wore  powers  of  ihe  s:imw  mimhorV 


332 


INVOLUTION. 


524.    To  find  the   Product  of  two  or   more   Powers  cT  the 
same  Number. 

22.  What  is  the  product  of  43  multiplied  by  42  • 
ANALYSIS. — 43 =4x4x4,  and  4.^=4x4;  therefore  in  the  product 
of  43  x  44, 4  is  taken  3  +  2,  or  5  times  as  a  factor.    But  3  and  2  are  the 
given  indices  ;  therefore  4  is  taken  as  many  times  as  a  factor  as  there 
are  units  in  the  indices.  Ans.  4".     Hence,  the 

EULE. — Add  the  indices,  and  the  sum  will  be  the  index 
of  the  product. 

23.  Mult,  23  by  22.  25.  Mult.  43  by  4*. 

24.  Mult  34  by  33.  26.  Mult.  5*  by  5'. 


FORMATION     OF     SQUARES. 

525.  To  find  the  Square  of  a   Number  in  the  Terms  of  its 

Parts. 

i.  Find  the  square  of  5  in  the  terms  of  the  parts  3  and  2. 

ANALYSIS. — Let  the  shaded  part  of  the 
diagram  represent  the  square  of  3  ; — each 
side  being  divided  into  3  inches,  its  con- 
tents are  equal  to  3  x  3,  or  9  sq.  in. 

The  question  now  is,  what  additions 
must  be  made  and  how  made,  to  preserve 
fhcform  of  this  square,  and  make  it  equal 
to  the  square  of  5. 

i  st.  To  preserve  the  form  of  the  square 
it  is  plain  equal  additions  must  be  made 
to  two  adjacent  sides ;  for,  if  made  on  one  side,  or  on  opposite  sides, 
the  figure  will  no  longer  be  a  square. 

2d.  Since  5  is  2  more  than  3,  it  follows  that  two  rows  of  3  squares 
each  must  be  added  at  the  top,  and  2  rows  on  one  of  the  adjacent 
sides,  to  make  its  length  and  brendth  each  equal  to  5.  Now  2  into  3 
plus  2  into  3  are  12  squares,  or  twice  the  product  of  the  two  parts 
2  and  3. 

But  the  diagram  wants  2  times  2  small  squares,  as  represented  by 
the  dotted  lines,  to  fill  the  corner  on  the  right,  and  2  times  2  or  4  is 
the  square  of  the  second  part.  We  have  then  9  (the  sq.  of  the  ist 
part),  12  (twice  the  prod,  of  the  two  parts  3  and  2),  and  4  (the  square 
of  the  zd  part).  But  9  +  12  +  4=25,  the  square  required. 


EVOLUTION.  333 

Again,  if  5  is  divided  into  4  and  i,  the  square  of  4  is  16,  twice  the 
prod,  of  4  into  i  is  8,  and  the  square  of  i  is  i.  But  16  +  8  +  1=25. 

2.  Eequired  the  square  of  25  in  the  terms  of  20  and  5. 

ANALYSIS. — Multiplying  20  by  20  gives  25  —    20  +  5 

400  (the  square  of  the  ist  part);   20x5  25        20  +  5 

plus  20  x  5  gives  200  (twice  the  prod,  of  ^2^     AOO  4-  100 

the  two  parts) ;  aud  5  into  5  gives  25  (the  Q3  . 

square  of  the  2d  part).    Now  400  +  200  +  25  -- — 

=625,  or  252.    Hence,  universally,  625  =400  +  200  +  25 

The  square  of  any  number  expressed  in  the  terms  of  its 
parts,  is  equal  to  the  square  of  the  first  part,  plus  twice  the 
product  of  the  two  parts,  plus  the  square  of  the  second  part. 

3.  What  is  the  square  of  23  in  the  parts  20  and  3  ? 

4.  What  is  the  sqrare  of  2%,  or  2  +•§  ?  Ans.  6£. 

5.  What  is  the  square  of  f  or  £  +  £?  Ans.  4. 


EVOLUTION. 

526.  Evolution  is  finding  a  root  of  a  number. 
A  Hoot  is  one  of  the  equal  factors  of  a  number. 

527.  Roots,  li-ke  powers,  are  divided  into  degrees;  as,  the 
square,  or  second  root;  the  cube,  or  third  root;  the  fourth 
root,  etc. 

528.  The  Square  Root  is  one  of  the  two  equal 
factors  of  a  number.    Thus,  5  x  5  =  25  ;  therefore,  5  is  the 
square  root  Of  25. 

529.  The    Cube  Hoot  is   one   of  the  three  equal 
factors  of  a  number.     Thus,  3x3x3  —  27;  therefore,  3  is 
the  cube  root  of  27,  etc. 


525.  To  what  is  the  square  of  a  number  equal  in  the  terms  of  its  parts? 
5*6.  What  is  evolution  f    A  root  1    528.  Square  root  ?    529.  Cube  root  ? 


334  EVOLUTION. 

530.  Roots  are  denoted  in  two  tuays:  ist.  By  prefixing 
to  the  number  the  character  (  \/),  called  the  radical  sign, 
with  a  figure  over  it  ;  as  V4,  V8. 

2d.  By  a  fractional  exponent  placed  above  the  number 
on  the  right.     Thus,  Vg,  or  9*j  denotes  the  sq.  root  of  9. 

NOTES.  —  i.  The  figure  over  the  radical  sign  (i/)  and  the  denomi- 
nator of  the  exponent  respectively,  denote  the  name  of  the  root. 

2.  In  expressing  th?  square  root,  it  is  customary  to  use  simply  the 
radical   sign  (i/),  the  2  being  understood.     Thus,  the   expression 
1/25=5,  is  read,  "  the  square  root  of  25  =  5." 

3.  The  term  radical  is  from  the  Latin  radix,  root.    The  sign  (j/) 
is  a  corruption  of  the  letter  r,  the  initial  of  radix. 

531.  A  Perfect  Power  is  a  number  whose  exact 
root  can  be  found. 

An  Imperfect  Power  is  a  number  whose  exact 
root  can  not  be  found. 

532.  A   Surd  is  the  root  of  an   imperfect  power. 
Thus,  5  is  an  imperfect  power,  and  its  square  root  2.23  + 
is  a  surd. 

NOTE.  —  All  roots  as  well  as  powers  of  r,  are  i. 
Eead  the  following  expressions  : 
i.  1/40.       3.  119*.        5.  1.5'.         7-  ^256.  9.  V££- 


2.  Vi5-       4-  2433.        6.  V29-        8.  '1/45-7-        1.        =   . 
1  1.  Express  the  cube  root  of  64  both  ways  ;  the  4th  root 
of  25  ;  the  jtli  root  of  81  ;  the  loth  root  of  100. 

533.  To  find   how   many  figures  the   Square  of  a   Number 
contains. 

ist.  Take  i  and  9,  the  least  and  greatest  integer  that  can 
be  expressed  by  one  figure;  also  10  and  99,  the  least  and 
greatest  that  can  be  expressed  by  two  integral  figures,  etc. 
Squaring  these  numbers,  we  have  for 

TheEoots:       i,     9,     10,       99,       100,         999,  etc. 

The  Sq-uares:  i,  81,  100,  9801,  10000,  998001,  etc. 

530.  How  are  roots  denoted?  531.  A  perfect  power?  Imperfect?  532.  A 
rord  ?  5«j3.  How  many  figures  has  the  square  of  a  number  ? 


SQUARE     ROOT.  335 

2(1.  Take  .  i  and  .9,  the  least  and  greatest  decimals  that 
can  be  expressed  by  one  figure;  also  .01  and  .99,  the  least 
and  greatest  that  can  be  expressed  by  two  decimal  figures, 
etc.  Squaring  these,  we  have  for 

The  Roots:     .1,    .9,    .01,      .99,      .001,        .999,       etc. 

The  Squares:  .01,  .81,  .0001,  .9801,  .000001,  .998001,  etc. 

By  inspecting  these  roots  and  squares,  we  discover  that 

Tlie  square  of  the  number  contains  twice  as  many  figures 
as  its  root,  or  twice  as  many  less  one. 

534.    To  find    how  many  figures   the  Square  Hoot   of  a 
Number  contain*. 

Divide  the  number  into  periods  of  two  figures  each, 
placing  a  dot  over  units'  place,  another  over  hundreds,  etc. 
The  root  will  have  as  many  figures  as  there  are  periods. 

REMARK. — Since  the  square  of  a  number  consisting  of  tens  and 
units,  is  equal  to  the  square  of  the  tens,  etc.,  when  a  number  has  two 
periods,  it  follows  that  the  left  hand  period  must  contain  the  squart 
of  the  tens  or  first  figure  of  the  root.  (Art.  525.) 


EXTRACTION     OF    THE     SQUARE    ROOT. 
535.  To  extract  the  Square  Root  of  a  Number. 

i.  A  man  wishes  to  lay  out  a  garden  in  the  form  of  a 
square,  which  shall  contain  625  sq.  yards :  what  will  be 
the  length  of  one  side  ? 

ANALYSIS. — Since  625  contains  two  periods,  its  root        OPEHATION. 
will  have  two  figures,  and  the  left  hand  period  con-  (etc 

tains  the  square  of  the  tens'  figure.     (Art.  534,  Hem.) 

But  the  greatest  square  of  6  is  4,  and  the  root  of  4 
is  2,  which  we  place  on  the  right  for  the  tens'  figure      45)225 
of  the  root.     Now  the  square  of  2  tens  or  20  is  400,  ?_?5_ 

and  625—400=225.    Hence.  225  is  twice  the  product  of 
the  tens'  figure  of  the  roct  into  the  units,  plus  the  square  of  the 
units.     (Art.  525.) 

534.  How  many  figures  has  the  square  root  of  a  number  ? 


336 


EXTRACTION  OF  SQUABE  BOOT. 


20  yds.         5  yds. 


But  one  of  the  factors  of  this  product  is  2  times  20  or  40 ;  there- 
fore the  other  factor  must  be  225  divided  by  40 ;  aud  225-7-40—5, 
the  factor  required.  (Art.  93.)  Taking  2  times  20  into  5=200 
(i.  e.,  twice  the  product  of  the  tens  into  the  units'  figure  of  the  root) 
from  225,  leaves  25  for  the  square  of  the  units'  figure,  the  square 
root  of  which  is  5  Hence  25  is  the  square  root  of  625,  and  is  there- 
fore the  length  of  one  side  of  the  garden. 

2d  ANALYSIS. — Let  the  shaded  part  of 
the  diagram  be  the  square  of  2  tens,  the 
first  figure  of  the  root ;  then  20  x  20,  or 
400  sq.  yds.,  will  be  its  contents.  Sub- 
tracting the  contents  from  the  given  area, 
we  have  625—400=225  sq.  yds.  to  be  added 
to  it.  To  preserve  its  form,  the  addition 
must  be  made  equally  to  two  adjacent 
sides.  The  question  now  is,  what  is  the 
width  of  the  addition. 


20  yds.         5  yds. 


Since  the  length  of  the  plot  is  20  yds.,  adding  a  strip  i  yard  wide 
to  two  sides  will  take  20  +  20  or  40  sq.  yds.  Now  if  40  sq.  yds.  will 
add  a  strip  i  yard  wide  to  the  plot,  225  sq.  yds.  -will  add  a  strip  as 
many  yds.  wide  as  40  is  contained  times  in  225  ;  and  40  is  contained 
in  225,  5  times  and  25  over. 

That  is,  since  the  addition  is  to  bo  mado  on  two  sides,  we  double 
the  root  or  length  of  one  side  for  a  trial  divisor,  and  find  it  is  con- 
tained in  225,  5  times,  which  Bhows  the  width  of  the  addition  to  be 
5  yards. 

Now  the  length  of  each  side  addition  being  20  yds.,  and  the  widtL 
5  yds.,  the  area  of  both  equals  20  x  5  +  20  x  5,  or  40  x  5  =  200  sq.  yards. 
But  there  is  a  vacancy  at  the  upper  corner  on  the  right,  whose  length 
and  breadth  are  5  yds.  each  ;  hence  its  area=5  x  5,  or  25  sq.  yards 
and  200  sq.  yd.  +  25  sq.  yd. =225  sq.  yd.  For  the  sake  of  finding  the 
area  of  the  two  side  additions  and  that  of  the  corner  at  the  same 
time,  we  place  the  quotient  5  on  the  right  of  the  root  already  found, 
and  also  on  the  right  of  the  trial  divisor  to  complete  it.  Multiplying 
the  divisor  thus  completed  by  5,  the  figure  last  placed  in  the  root, 
we  have  45x5  =  225  sq.  yds.  Subtracting  this  product  from  the 
•lividend,  nothing  remains.  Therefore,  etc.  Hence,  the 


535.  What  is  the  first  step  in  extracting  the  square  root  of  a  numher  ?  The 
second?  Third?  Fourth?  How  proved?  Note.  If  the  trial  divisor  is  not  con- 
tained in  the  dividend,  what  must  be  done?  if  there  is  a  remainder  after  the 
root  of  the  la<t  period  is  found,  what?  now  many  decimals  docs  the  root  of  a 
decimal  fraction  have  ? 


EXTRACTION    OF    SQUARE    BOOT.  33? 

KULE. — I.  Divide  the  number  into  periods  of  two  figures 
each,  putting  a  dot  over  units,  then  over  every  second  figure 
towards  the  left  in  whole  numbers,  and  towards  the  right 
in  decimals. 

II.  Find  the  greatest  square  in  the  left  hand  period,  and 
place  its  root  on  the  right.    Subtract  this  square  from  the 
period,  and  to  the  right  of  the  remainder  bring  down  the 
next  period  for  a  dividend. 

III.  Double  the  part  of  the  root  thus  found  for  a  tria't 
divisor;  and  finding  how  many  times  it  is  contained  in  the 
dividend,  excepting  the  right  hand  figure*  annex  the  quotient 
both  to  the  root  and  to  the  divisor. 

IV.  Multiply   tJie  divisor  thus  increased  by  this   last 
figure  placed  in  the  root,  subtract  the  product,  and  bring 
down  the  next  period. 

V.  Double  the  right  hand  figure  of  the  last  divisor,  and 
proceed  as  before,  till  the  root  of  all  the  periods  is  found. 

PROOF. — Multiply  the  root  into  itself.    (Art  528.) 

NOTES. — i.  If  the  trial  divisor  is  not  contained  in  the  dividend, 
annex  a  cipher  both  to  the  root  and  to  the  divisor,  and  bring  down 
the  next  period. 

— -2r  Since  the  product  of  the  trial  divisor  into  the  quotient  figure 
cannot  exceed  the  dividend,  allowance  must  be  made  for  carrying,  if 
the  product  of  this  figure  into  itself  exceeds  9. 

. — -ylt  sometimes  happens  that  the  remainder  is  larger  than  the 
divisor ;  but  it  does  not  necessarily  follow  from  this  that  the  figure 
in  the  root  is  too  small,  as  in  simple  division. 

4.  If  there  is  a  remainder  after  the  root  of  the  last  period  is  found, 
annex  periods  of  ciphers,  and  proceed  as  before.     The  figures  of  the 
root  thus  obtained  will  be  decimals. 

5.  The  square  root  of  a  decimal  fraction  is  found  in  the  same  way 
as  that  of  a  whole  number;  and  the  root  will  have  as  many  decimal 
figures  as  there  are  periods  of  decinu  1;  in  the  given  number. 

6.  The  left  hand  period  in  whole  numbers  may  have  but  one  figure; 
luit   in   decimals,  each  period  must   have   two  figures.     (Art.  533.} 
Hence,  if  the  number  lias  but  one  decimal  figure,  or  an  odd  number 
of  decimals,  a  cipher  must  be  annexed  to  complete  the  period. 

536.   REASONS. — I.   Dividing  the  number  into  periods  of  two 
figures  each,  shows  how  many  figures  the   root  will  contain,  and 
15 


338  EXTRACTION     OF     SQUARE     ROOT. 

enables  us  to  find  its  first  figure.  For,  the  left  hand  period  contains 
the  square  of  this  figure,  and  from  the  square  the  root  is  easily 
found.  (Art.  534,  Bern.) 

2.  Subtracting  the  square  of  the  first  figure  of  the  root  from  the 
left  hand  period,  shows  what  is  left  for  the  other  figures  of  the  root. 

3.  The  object  of  doubling  the  first  figure  of  the  root,  and  dividing 
the  remainder  by  it  as  a  trial  divisor,  is  to  find  the  next  figure  of  the 
root.     The  remainder  contains  tioice  th-3  product  of  the  tens  into  the 
units  ;  consequently,  dividing  this  product  by  double  the  tens'  factor, 
the  quotient  will  be  the  other  factor  or  units'  figure  of  the  root. 

4.  Or,  referring  to  the  diagram,  it  is  doubled  because  the  remainder 
must  be  added  to  two  sides,  to  preserve  iheform  of  the  square. 

5.  The  right  hand  figure  of  the  dividend  is  excepted,  to  counter- 
balance the  omission  of  the  cipher,  which  properly  belongs  on  tho 
right  of  the  trial  divisor. 

6.  The  quotient  figure  is  placed  in  the  root  ;  it  is  also  annexed  to 
the  trial   divisor  to  complete  it.     The  divisor   thus  completed   is 
multiplied  by  the  second  figure  of  the  root  to  find  the  contents  of  the 
additions  thus  made. 

The  reasons  for  the  steps  in  obtaining  other  figures  of  the  root 
may  bo  shown  in  a  similar  manner. 


2.  What  is  the  square  root  of  381.0304?        .4«s.  19.52. 

SUGGESTION.  —  Since  the  number  contains  decimals,  we  begin  at 
the  units'  place,  and  counting  both  ways,  have  four  periods  ;  as, 
381.0304.  The  root  will  therefore  have  4  figures.  But  there  are  two 
periods  of  decimals  ;  hence  we  point  off  two  decimals  in  the  root. 

3.  What  is  the  square  root  of  1012036  ?  Ans.  1006. 

4.  What  is  the  square  root  of  2  ?  Ans.  1.41421  +. 

Extract  the  square  root  of  the  following  numbers  : 


5.  182329. 

ii.  .1681. 

17-5- 

23-  I9-5364- 

6.  516961. 

12.  .725. 

18.  7. 

24.  3283.29. 

7-  595984- 

13.   .I26l. 

19.  8. 

25.  87.65. 

8.  3.580. 

14.  2.6752. 

20.  10. 

26.  123456789. 

9.  .4096. 

15.  4826.75. 

21.  II. 

27.  61723020.96. 

10.  .120409. 

1  6.  452-634. 

22.  12. 

28.  9754.60423716. 

536.  Why  divide  the  number  into  periods  of  two  figures  each  ?  Why  subtract 
the  square  of  the  first  figure  from  the  period  ?  Why  double  the  firs»t  figure  of  the 
root  for  a  trial  divisor?  Why  omit  the  right  hand  figure  of  the  dividend?  Why 
place  the  quotient  figure  on  the  right  of  the  trial  divisor?  Why  multiply  the 
trial  divisor  thus  completed  by  the  figure  last  placed  in  the  root? 


APPLICATIONS    OF   SQUARE     ROOT.         339 

537.  To  find  the  Square  Root  of  a  Common  Fraction. 

I.  When  the  numerator  and  denominator  are  both  perfect 
squares,  or  can  be  reduced  to  such,  extract  the  square  root 
of  each  term  separately. 

II.  When  they  are  imperfect  squares,  reduce  them  to 
decimals,  and  proceed  as  above. 

NOTE. — To  find  the  square  root  of  a  mixed  number,  reduce  it  to 
an  improper  fraction,  and  proceed  as  before. 

29.  "What  is  the  square  root  of  T\? 
ANALYSIS. — -^—^,  and  -1/^=5,  Ans. 

30.  "What  is  the  square  root  off?  .^^..7745  +, 

31.  "What  is  the  square  root  of  if?  Ans.  1.1726+. 

Find  the  square  root  of  the  following  fractions : 

32-  iff-  35-  6f  38-  Iff-  41-  !•&• 

33.  f.  36.  13$.  39.  f£§.  42.  27^. 

34- A-  37-  i7f-  4o.  HI-  43- 


APPLICATIONS. 

538.    To  find  the  Side  of  a  Square  equal  in  area  to  a 
given  Surface. 

1.  Find  the  side  of  a  square  farm  containing  40  acres. 

ANALYSIS. — In  i  acre  there  are  160 

,   .  ,.  OPERATION. 

eq.   rods,   and   m  40  acres,   40  times 

!6o,  or  6400  sq.  rods.    The   ^6400=     4°  A.  x  160  =  6400  Sq.  r. 
80  r.    Therefore  the  side  of  the  farm  is      4/6400  =  80  r.  Ans. 
80  linear  rods.     Hence,  the 

RULE. — I.  Extract  the  square  root  of  the  given  surface. 

NOTE. — The  root  is  in  linear  units  of  the  same  name  as  the  given 
surface. 

2.  "What  is  the  side  of  a  square  tract  of  laud  containing 
1 102  acres  80  sq.  rods? 

537.  How  find  the  square  root  of  a  common  fraction?    Note.  Of  a  mixed 
number  ?    538.  How  find  the  side  of  a  square  equal  to  a  given  surface  f 


340         APPLICATIONS    OF    SQUAfiE     KOOT. 


3.  How  many  rods  of  fencing  does  it  require  to  inclose 
a  square  farm  which  contains  122  acres  30  sq.  rods  ? 

4.  A  bought  14161  fruit  trees,  which  he  planted  so  as 
to  form  a  square :  how  many  trees  did  he  put  in  a  row  ? 

5.  A  general  has  an  army  of  56644  men:  how  many 
must  he  place  in  rank  and  file  to  form  them  into  a  square  'J. 


539.   A  Triangle  is  a  figure  having  three 
sides  and  three  angles. 

A  Hiyht-angled  Triangle  is  one  that 
contains  a  right  angle.  (Art.  260.) 

The  side  opposite  the  right  angle  is  called  the 
hypothenuse;  the  other  two  sides  the  base  and 
perpendicular. 


Base 


540.  The  square  described  on  the  hypothenuse  of  a  right- 
angled  triangle  is  equal  to  the  sum  of  the  squares  described 
on  the  other  two  sides.* 


541.  The  truth  of  this  principle  may 
be  illustrated  thus :  Take  any  right-angled 
triangle  ABC;  let  the  hypothenuse  h, 
be  5  in.,  the  base  b,  4  in.,  and  the  per- 
pendicular p,  3  in.  It  will  be  seen  that  the 
square  of  h  contains  25  sq.  in.,  the  square 
of  b  16  sq.  in.,  and  the  square  of  p  9  sq.  in. 
Now  25  — 16  +  9,  which  accords  with  the 
proposition.  In  like  manner  it  may  be 
shown  that  the  principle  is  true  of  all 
right-angled  triangles.  Hence, 


542.  To  find  the  Hypothenuse,  the  Base  and  Perpendicular 
being  given. 

To  the  square  of  the  base  add  the  square  of  the  per- 
pendicular, and  extract  the  square  root  of  their  sum. 

539.  What  is  a  triangle  1  A  right-angled  triangle  ?  Draw  a  right-nngled 
triangle?  The  side  opposite  the  right-angle  called?  The  other  two  sides? 
540.  To  what  is  the  square  of  the  hypothennse  equal?  541.  Illustrate  this  prin- 
ciple by  a  figure?  542.  How  find  the  hypothenuse  when  the  base  and  per- 
pendicular are  given  ? 

*  Thomson's  Geometry,  IV,  1 1 ;  Euclid,  I,  47. 


APPLICATIONS     OF     SQUARE    ROOT.         341 

543.  To  find  the  Base,  the   Hypothenuse  and   Perpendicular 

being  given. 

From  the  square  of  the  hypothenuse  take  the  square 
of  the  perpendicular,  and  extract  the  square  root  of  the 
remainder. 

544.  To   find    the  Perpendicular,  the   Hypothenuse   and 

Base  being  given. 

From  the  square  of  the  hypothenuse  take  the  square 
of  the  base,  and  extract  the  square  root  of  the  remainder. 

NOTE. — The  pupil  should  draw  figures  corresponding  with  the 
conditions  of  the  following  problems,  and  indicate  the  parts  given : 

6.  The  perpendicular  height  of  a  flag-staff  is  36  ft: 
what  length  of  line  is  required  to  reach  from  its  top  to  a 
point  in  a  level  surface  48  ft.  from  its  base  ? 

SOLUTION. — The  square  of  the  base    ..  =48x48=2304 
"        "        perpendicular = 36  x  36=1296 
The  square  root  of  their  sum =1/3600 =60  ft.  Ans. 

7.  The  hypothenuse  of  a  right-angled  triangle  is  135  yds., 
the  perpendicular  81  yds. :  what  is  the  base  ? 

8.  One  side  of  a  rectangular  field  is  40  rods,  and  the 
distance  between  its  opposite  corners  50  rods:  what  is  the 
length  of  the  other  side  ? 

9.  Two  vessels   sail  fronv  the  same  point,  one  going 
due  south  360  miles,  the  other  due  east  250  miles:  how 
far  apart  were  they  then  ? 

10.  The  height  of  a  tree  on  the  bank  of  a  river  is  100  ft., 
and  a  line  stretching  from  its  top  to  the  opposite  side  is 
144  ft. :  what  is  the  width  of  the  river  ? 

n.  The  side  of  a  square  room  is  40  feet:  what  is  the 
distance  between  its  opposite  corners  on  the  floor  ? 

12.  A  tree  was  broken  35  feet  from  its  root,  and  struck 
the  ground  21  ft.  from  its  base:  what  was  the  height  of 
the  tree  ? 

543.  How  find  the  base  when  the  other  two  sides  are  given?  544.  How  find 
the  perpendicular  when  the  other  two  sides  are  given  ? 


342  SIMILAE     FIGURES. 

SIMILAR    FIGURES. 

545.  Similar  Figures  are  those  which  have  the 
same  form,  and  their  like  dimensions  proportional. 

NOTES. — i.  All  circles,  of  whatever  magnitude,  are  similar. 

2.  All  squares,  equilateral  triangles,  and  regular  polygons  are 
similar.     And,  universally, 

All  rectilinear  figures  are  similar,  when  their  several  angles  are 
equal  each  to  each,  and  their  like  dimensions  proportional. 

3.  The  like  dimensions  of  circles  are  their  diameters,  radii,  and 
circumferences. 

546.  The  areas  of  similar  figures  are  to  each  other  as 
the  squares  of  their  like  dimensions.     And, 

Conversely,  the  like  dimensions  of  similar  figures  are  to 
each  other  as  the  square  roots  of  their  areas. 

13.  If  one  side  of  a  triangle  is  12  rods,  and  its  area 
72  sq.  rods,  what  is  the  area  of  a  similar  triangle,  the 
corresponding  side  of  which  is  8  rods  ? 

SOLUTION. — (12)* :  (8)2 : :  72  :  Ans.,  or  32  sq.  rods. 

14.  If  one  side  of  a  triangle  containing  36  sq.  rods  is 
8  rods,  what  is  the  length  of  a  corresponding  side  of  a 
similar  triangle  which  contains  81  sq.  rods? 

SOLUTION. — ^36  :  j/8i : :  8  :  Ans.,  or  12  rods. 

15.  If  a  pipe  2  inches  in  diameter  will  fill  a  cistern  in 
42  min.,  in  what  time  will  a  pipe  7  in.  in  diameter  fill  it  ? 

547.  To  find  a  Mean  Proportional  between  two  Numbers. 

1 6.  What  is  the  mean  proportional  between  4  and  16? 
ANALYSIS. — When  three  numbers  are  proportional,  the  product  of 

the  extremes  is  equal  to  the  square  of  the  mean.    (Arts.  487,  490.) 
But  4x  16=64 ;  an(i  1/64=8.  Ans.    Hence,  the 

RULE. — Extract  the  square  root  of  their  product. 

Find  the  mean  proportional  between  the  following : 
17.  4  and  36.  19.  56  and  72.  21.  $$  and  -^-. 

*8.  36  and  81.          20.  .49  and  6.25.          22.  -^  and 


545.  What  are  similar  figures  ?    546.  How  arc  their  areas  to  each  other  Y 


FOKMATION     OF     CUBES.  343 

FORMATION    OF     CUBES. 

548.  To  find  the  Cube  of  a  Number  in  the  terms  of  its  parts, 
i.  FiaJ  the  cube  of  32  in  the  terms  of  the  parts  30  and  2. 

ANALYSIS.— 32=30  +2,  and  323=( 

32=30  +2 
"96"=: 
64= 


IO24=302  +  2  x  V3O  x  2)  +  2s 

32  =  3O  +  2  

3072  = 303  +  2  x  (30s  x  2)  4-  (30  x  22) 

2048  =  (30*  X  2)  +  2  X  v30j<  22)  +  28 

32768=30^3  X  (302  X  2)  +  3  X  (30  X  Z9)  +  23. 

Or  thus :  Let  the  diagram  represent  a 
cube  whose  side  is  30  ft. ;  its  contents^ 
the  cube  of  3  tens,  or  2700  cu.  ft. 

What  additions  must  be  made  to  this 
cube,  and  how  made,  to  preserve  its  form, 
and  make  it  equal  to  the  cube  of  32. 

ist.  To  preserve  its  cubical  form,  the 
additions  must  be  equally  made  on  three 
adjacent  sides ;  as  the  top,  front,  and  right. 

2d.  Since  32  is  2  more  than  30,  it  follows  that  this  cube  must  be 
made  2  ft.  longer,  2  ft.  wider,  and  2  ft.  higher,  that  its  length, 
breadth,  and  thickness  may  each  be  32  ft. 

But  as  the  side  of  this  cube  is  30  ft.,  the  contents  of  each  of  these 
additions  must  be  equal  to  the  square  of  the  tens  (3O2)  into  2,  ihe  units, 
and  their  sum  must  be  3  times  (3O2  x  2)=3  x  900  x  2=5400  cu.  ft. 

But  there  are  three  vacancies  along  the  edges  of  the  cube  adjacent 
to  the  additions.  Each  of  these  vacancies  is  30  ft.  long,  2  ft.  wide, 
and  2  ft.  thick  ;  hence,  the  contents  of  each  equals  30  x  22,  and  the  sum 
of  their  contents  equals  3  times  the  tens  into  the  square  of  the  units 
=3  times  (30  x  22)=3  x  30  x  4=360  cu.  ft.  But  there  is  still  another 
vacancy  at  the  junction  of  the  corner  additions,  whose  length, 
breadth,  and  thickness  are  each  2  ft.,  and  whose  contents  are  equal 
to  23=2x  2x2  =  8.  The  cube  is  now  complete.  Therefore  32*= 
27000  (3os)  +  54Oo  (3  times  3O2x2)  +  3&o  (3  times  30x2*) +  8  (23)= 
32768  cu.  ft.  In  like  manner  it  may  be  shown  that, 

548.  To  what  is  the  cube  of  a  number  consisting  of  tens  and  units  equal? 


344  FORMATION     OF     CUBES. 

The  cube  of  any  number  consisting  of  tens  and  units  is 
equal  to  the  cube  of  the  tens,  plus  3  times  the  square  of  tlw 
tens  into  the  units,  plus  3  times  the  tens  into  the  square  of 
the  units,  plus  the  cube  of  the  units. 

549.  To  find  how  many  figures  the  Cube  of  a  Number  contains. 

ist.  Take  i  and  9,  also  10  and  99,  100  and  999,  etc.,  the  least  and 

greatest  integers  that  can  be  expressed  by  one,  two,  three,  etc.,  figures. 

2d.  In  like  manner  take  .1  and  .9,  also  .01  and  .99,  etc.,  the  least 

and  greatest  decimals  that  can  be  expressed  by  one,  two,  etc.,  decimal 

figures.    Cubing  these,  we  have 

The  root    land     r'=i,      -     -     -     -       .1  and     .I3=.i 
"  9    "        93--729,       -     -    -       .9    "        .g3=.729 

"  IO     "         IO3=IOOO,       -      -      -       .OI      "        .OI3=.OOOOOI 

99    "      995=970299,      -    -     .99    "      .gg3-. 970299 
"        100    "    ioo3=ioooooo,    -     -  .001    "    .ooi3 =.000000001 
999    "    9993= 997002999,      -  .999    "    -9993=. 997002999. 
By  comparing  these  roots  and  their  cubes,  we  discover  that 
TJie  cube  of  a  number  cannot  have  more  than  three  times 
as  many  figures  as  its  root,  nor  but  two  less.    Hence, 

550.  To  find  how  many  figures  the  Cube  Hoof,  contains. 

Divide  the  number  into  periods  of  three  figures  each, 
putting  a  dot  over  units,  then  over  every  third  figure 
towards  the  left  in  whole  numbers,  and  towards  the  right 
in  decimals, 

REMAKKS. — i.  Since  the  cube  of  a  number  consisting  of  tens  and 
units  is  equal  to  the  cube  of  the  tens,  plus  3  times  the  square  of  the 
tens  into  the  units,  etc.,  when  a  number  has  two  periods,  it  follows 
that  the  left  hand  period  must  contain  the  cube  of  the  tens,  or  first 
figure  of  the  root. 

2.  The  left  hand  period  in  whole  numbers  may  be  incomplete, 
having  only  one  or  two  figures ;  but  in  decimals  each  period  must 
always  have  three  figures.  Hence,  if  the  decimal  figures  in  a  given 
number  are  less  than  three,  annex  cipher*  to  complete  the  period. 

How  many  figures  in  the  cube  root  of  the  following : 
2.  340566.  4.  576.453.  6.  32.7561. 

3-  1467-  5-  5-732I-  7-  456785- 

549.  How  many  figures  has  the  cube  of  a  number? 


EXTRACTION     OF    THE     CUBE     BOOT.         345 

EXTRACTION     OF     THE     CUBE     ROOT. 
551.  To  extract  the  Cube  Root  of  a  Number. 

i.  A  man  having  32768  marble  blocks,  each  being  a 
cubic  foot,  wishes  to  arrange  them  into  a  single  cube: 
what  must  be  its  side  ? 

ANALYSIS.  —  Since   32768  contains  two   periods,  OPERATION. 

we  know  its  root  will  have  two  figures;  also  that  12768(32 

the  left  hand  period  contains  the  cube  of  the  tens  2 
or  first  figure  of  the  root.    (Art.  550,  Hem.) 


5768 


5768 


,         . 

The  greatest  cube  in  32  is  27,  and  its  root  is  3,        '  ~ 
which  we  place  on  the  right  for  the  tens  or  first 
figure  of  the  root.     Subtracting  its  cube  from  the 
first  period,  and  bringing  down  the  next  period 
to  the  right  of  the  remainder,  we  have  5768,  which 
by  the  formation  of  the  cube  is  equal  to  3  times  the  square  of  the 
tens'  figure  into  the  units,  plus  3  times  the  tens  into  the  square  of 
the  units,  plus  the  cube  of  the  units.     We  therefore  place  3  times 
the  square  of  the  tens  or  2700,  on  the  left  of  the  dividend  for  a 
trial  divisor  ;  and  dividing,  place  the  quotient  2  on  the  right  for  the 
units'  figure  of  the  root. 

To  complete  the  divisor,  we  add  to  it  3  times  30  the  tens  into  2 
units=i8o;  also  4,  the  square  of  the  units,  making  it  2884.  Multi- 
plying the  divisor  thus  completed  by  the  units'  figure  2,  we  have 
2884  x  2  =  5768,  the  same  as  the  dividend.  Ans.  32  ft. 

Or  thus  :  Let  the  cube  of  30,  the  tens  of  the  root,  be  represented 
by  the  large  cube  in  the  set  of  cubical  blocks.*  The  remainder  5768, 
is  to  be  added  equally  to  three  adjacent  sides  of  this  cube. 

To  ascertain  the  thickness  of  these  side  additions,  we  form  a  trial 
divisor  by  squaring  3,  tho  first  figure  of  the  root,  with  a  cipher 
annexed,  for  the  area  of  one  side  of  this  cube,  and  multiply  this 
square  by  3  for  the  three  side  additions.  Now  3O2=  30  x  30=900; 
and  900x3  =  2700,  the  trial  divisor.  Dividing  5768  by  2700,  the 
quotient  is  2,  which  shows  that  the  side  additions  are  to  be  2  ft. 
thick,  and  is  placed  on  the  right  for  the  units'  figure  of  the  root. 

551.  The  first  step  in  extracting  the  cube  root?  The  second?  Third? 
Fourth  1  Fifth  ?  Note.  If  the  trial  divisor  is  not  contained  in  the  dividend,  how 
proceed  ?  If  there  is  a  remainder  after  the  root  of  the  last  period  is  found,  how  ? 


*  Every  school  in  which  the  cuhe  root  is  taught  is  presumed  to  be  furnished 
with  a  set  of  Cubical  Blocks, 


346         EXTRACTION     OF     THE      CUBE     BOOT. 

To  represent  these  additions,  place  the  corresponding  layers  on 
the  top,  front,  and  right  of  the  large  cube.  But  we  discover  three 
vacancies  along  the  edges  of  the  large  cube,  each  of  which  is  30  ft. 
long,  2  ft.  wide,  and  2  ft.  thick.  Filling  these  vacancies  with  the 
corresponding  rectangular  blocks,  we  discover  another  vacancy  at 
the  junction  of  the  corners  just  filled,  whose  length,  breadth,  and 
thickness  are  each  2  ft.  This  is  filled  by  the  small  cube. 

To  complete  the  trial  divisor,  we  add  the  area  of  one  side  of  each 
of  the  corner  additions,  viz.,  30  x  2  x  3,  or  180  sq.  ft.,  also  the  area  of 
one  side  of  the  small  cube=2  x  2,  or  4  sq.  ft.  Now  2700+  180+4= 
2884.  The  divisor  is  now  composed  of  the  area  of  3  sides  of  the 
large  cube,  plus  the  area  of  one  side  of  each  of  the  corner  additions, 
plus  the  area  of  one  side  of  the  small  cube,  and  is  complete. 

Finally,  to  ascertain  the  contents  of  the  several  additions,  we 
multiply  the  divisor  thus  completed  by  2,  the  last  figure  of  the  root ; 
and  2884  x  2=5768.  (Art.  249.)  Subtracting  the  product  from  the 
dividend,  nothing  remains.  Hence,  the 

KULE.— I.  Divide  the  number  into  periods  of  three  figures 
each,  putting  a  dot  over  units,  then  over  every  third  figure 
towards  the  left  in  ivhole  numbers,  and  towards  the  right 
in  decimals. 

II.  Find  the  greatest  cube  in  the  left  hand  period,  and 
place  its  root  on  the  right.     Subtract  its  cube  from  the 
period,  and  to  the  right  of  the  remainder  bring  down  the 
next  period  for  a  dividend. 

III.  Multiply  the  square  of  the  root  thus  found  with  a 
cipher  annexed,  by  three,  for  a  trial  divisor;  and  finding 
hoiv  many  times  it  is  contained  in  the  dividend,  icrite  the 
quotient  for  the  second  figure  of  the  root. 

IV.  To  complete  the  trial  divisor,  add  to  it  three  times 
the  product  of  the  root  previously  found  witli  a  cipher 
annexed,  into  the  second  root  figure,  also  the  square  of  the 
second  root  figure. 

V.  Multiply  the  divisor  thus  completed  by  the  last  figure 
placed  in  the  root.     Subtract  the  product  from  the  divi- 
dend; and  to  the  right  of  the  remainder  bring  down  the 
next  period  for  a  new  dividend.     Find  a  new  trial  divisor, 
as  before,  and  thus  proceed  till  the  root  of  the  last  period  is 
found. 


EXTRACTION     OF    THE     CUBE     BOOT.        347 

NOTES. — i.  If  a  trial  divisor  is  not  contained  in  the  dividend,  put 
a  cipher  in  the  root,  two  ciphers  on  the  right  of  the  divisor,  and  bring 
down  the  next  period. 

2.  If  the  product  of  the  divisor  completed  into  the  figure  last 
placed  in  the  root  exceeds  the  dividend,  the  root  figure  is  too  large. 
Sometimes  the  remainder  is  larger  than  the  divisor  completed ;  but 
it  does  not  necessarily  follow  that  the  root  figure  is  too  small. 

3.  When  there  are  three  or  more  periods  in  the  given  number,  the 
first,  second,  and  subsequent  trial  divisors  are  found  in  the  same 
manner  as  when  there  are  only  two.     That  is,  disregarding  its  true 
local  value,  wo  simply  multiply  the  square  of  the  root  already  found 
with  a  cipher  annexed,  by  3,  etc. 

4.  If  there  IP-  a  remainder  after  the  root  of  the  last  period  is  found, 
annex  periods  of  ciphers,  and  pioceed  as  before.     The  root  figures 
thus  obtained  will  be  decimals. 

5.  The  cube  root  of  a  decimal  fraction  is  found  in  the  same  way  as 
that  of  a  whole  number ;  and  will  have  as  many  decimal  figures  as 
there  are  periods  of  decimals  in  the  number.     (Art.  549.) 

552.  REASONS. — i.  Dividing  the  number  into  periods  shows 
how  many  figures  the  root  contains,  and  enables  us  to  find  the  first 
figure  of  the  root.  For,  the  left  hand  period  contains  the  cube  of 
the  first  figure  of  the  root.  (Art.  550.) 

2.  The  object  of  the  trial  divisor  is  to  find  the  next  figure  of  the 
root,  or  the  thickness  of  the  side  additions.     The  root  is  squared  to 
find  the  area  of  one  side  of  the  cube  whose  root  is  found,  the  cipher 
being  annexed  because  the  first  figure  is  tens.     This  square  is  multi- 
plied by  3,  because  the  additions  are  to  be  made  to  three  sides. 

3.  The  root  previously  found  is  multiplied  by  this  second  figure 
to  find  the  area  of  a  side  of  one  of  the  vacancies  along  the  edges  of 
the  cube  already  found.     This  product  is  multiplied  by  3,  because 
there  are  three  of  these  vacancies ;  and  the  product  is  placed  under 
the  trial  divisor  as  a  correction.     The  object  of  squaring  the  second 
figure  of  the  root  is  to  find  the  area  of  one  side  of  the  cubical  vacancy 
at  the  junction  of  the  corner  vacancies,  and  with  the  other  correction 
this  is  added  to  the  trial  divisor  to  complete  it. 

4.  The  divisor  thus  completed  is  multiplied  by  the  second  figuro 
of  the  root  to  find  the  contents  of  the  several  additions  now  made. 


552.  Why  divide  the  number  into  periods  of  three  figures  ?  What  is  the  object 
of  a  trial  divisor  ?  Why  square  the  root  already  found  ?  Why  annex  n  cipher  to 
it?  Why  multiply  this  square  by  3?  Why  is  the  root  previously  found  multi- 
plied by  the  second  figure  of  the  root?  Why  multiply  this  product  by  3*  Why 
square  the  second  figure  of  the  root  ?  Why  multiply  the  divisor  wh«n  completed 
by  the  second  figure  of  the  root* 


348        EXTRACTION     OF    THE     CUBE     ROOT. 

2.  "What  is  the  cube  root  of  130241.3  ? 

ANALYSIS.  —  Having  completed   the  OPEEATION. 

period  of  decimals   by  annexing  two  130241.300(50.64 

ciphers,  we  find  the  first  figure  of  the  I2!. 

root  as  above.      Bringing  down   the 
nest    period,   the    dividend    is    5241.      75OOO°  5241-3°' 
The  trial  divisor  7500  is  not  contained 
in  the  dividend;  therefore  placing  a     _  3_ 
cipher  in  the   root  and  two  on  the     759°3^  455421^ 
right  of  the  divisor,  we  bring  down  687084  Rem. 

the  next  period,  and  proceed  as  before. 

3.  Cube  root  of  614125  ?  6.  Cube  root  of  3  ? 

4.  Cube  root  of  84.604  ?  7.  Cube  root  of  21.024576  ? 

5.  Cube  root  of  373248  ?  8.  Cube  root  of  17  ? 
9.  "What  is  the  cube  root  of  705919947264? 

10.  What  is  the  cube  root  of  .253395799  ? 

11.  What  is  the  side  of  a  cube  which  contains  628568 
cu.  yards  ? 

NOTE.  —  The  root  is  in  linear  units  of  the  same  name  as  the  given 
contents. 

12.  What  is  the  side  of  a  cube  equal  to  a  pile  of  wood 
40  ft.  long,  1  5  ft.  wide,  and  6  ft.  high  ? 

13.  What  is  the  side  of  a  cubical  bin  which  will  hold 
1000  bu.  of  corn;  allowing  2150.4  cu.  in.  to  a  bushel  ? 

553.  To  find  the  Cube  Root  of  a  Common  Fraction. 

If  the  numerator  and  denominator  are  perfect  cubes,  of 
can  be  reduced  to  such,  extract  the  cube  root  of  each. 

Or,  reduce  the  fraction  to  a  decimal,  and  proceed  as  before 
NOTE.  —  Reduce  mixed  numbers  to  improper  fractions,  etc. 

14.  What  is  the  cube  root  of  -f^s  ?  Am.  f. 

15.  Cube  root  off?  Ans.  .75+. 
1  6.  Cube  root  of  jf&  ?            18.  Cube  root  of  TV2\  ? 
17.  Cube  root  of  -ff^fj  ?          19.  Cube  root  of  8  if  ? 


553.  How  find  the  cube  root  of  a  common  fraction?    Note.  Of  a  mixed  number? 


SIMILAR     SOLIDS.  349 


APPLICATIONS. 

554.  Similar  Solids  are  those  which  have  the  same 
form,  and  their  like  dimensions  proportional. 

NOTES. — i.  The  like  dimensions  of  spheres  are  their  diameters, 
radii,  and  circumferences;  those  of  cubes  are  their  sides. 

2.  The  like  dimensions  of  cylinders  and  cones  are  their  altitudes 
and  the  diameters  or  the  circumferences  of  their  bases. 

3.  Pyramids  are  similar,  when  their  bases  are  similar  polygons 
and  their  altitudes  proportional. 

4.  Polyhedrons  (i.  e.,  solids  included  by  any  number  of  plane  faces) 
are  similar,  when  they  are  contained  by  the  same  number  ot  similai 
polygons,  and  all  their  solid  angles  are  equal  each,  to  each. 

555.  The  contents  of  similar  solids  are  to  each  other  aa 
the  cubes  of  their  like  dimensions;  and, 

Conversely,  the  like  dimensions  of  similar  solids  are  a& 
the  cube  roots  of  their  contents. 

1.  If  the  side  of  a  cubical  cistern  containing  1728  en.  in. 
is  1 2  in.,  what  arc  the  contents  of  a  similar  cistern  whose 
side  is  2  ft.  Ans.  i 3  :  23  : :  1728  cu.  in. :  con.,  or  13824  en.  in. 

2.  If  the  side  of  a  certain  mound  containing  74088  cu.  ft. 
is  84  ft.,  what  is  the  side  of  a  similar  mound  which  con- 
tains 17576  en.  ft? 

3.  If  a  globe  4  in.  in  diameter  weighs  9  Ibs ,  what  is  the 
weight  ot  a  globe  8  in.  in  diameter?  Ans.  72  Ibs 

4.  If  8  cubic  piles  of  wood,  the  side  of  each  being  8  ft-., 
were  consolidated  into  one  cubic  pile,  what  would  be  the 
length  of  its  side  ? 

5.  If  a  pyramid  60  ft.  high  contains  12500  cu.  ft.,  what 
are  the  contents  of  a  similar  pyramid  whose  height  is  20  ft.? 

6.  If  a  conical  stack  of  hay  15  ft.  high  contains  6  tons, 
what  is  the  weight  of  a  similar  stuck  whose  height  is  12  ft.? 

554.  What  are  similar  s.  lii--?     .'u/'V.  \Vlmt  .ire  like  dimensions  of  spheres  t 
Of  cylinders  and  cones  f    Of  pyramids  ?    555.  What  relation  have  similar  solids  ? 


ARITHMETICAL    PROGRESSION. 

556.  An  Arithmetical  Progression  is  a  series 
of  numbers  which  increase  or  decrease  by  a  common  dif- 
ference. 

NOTE. — The  numbers  forming  the  series  are  called  the  Terms. 
The  first  and.  last  terms  are  the  Extremes;  the  intermediate  terms 
the  Means.  (Arts.  476,  489.) 

557.  The  Common  Difference  is  the  difference 
between  the  successive  terms. 

558.  An  Ascending  Series  is  one  in  which  the  successive 
terms  increase ;  as,  2,  4,  6,  8, 10,  etc.,  the  common  difference  being  2. 

A  Descending  Series  is  one  in  which  the  successive  terms 
decrease ;  as,  15,  12,  9,  6,  etc.,  the  common  difference  being  3. 

559.  In  arithmetical  progression  there  are  five  parts  or  elements 
to  be  considered,  viz. :  the  first  term,  the  last  term,  the  number  of 
terms,  the  common  difference,  and  the  sum  of  all  the  terms.    These 
parts  are  so  related  to  each  other,  that  if  any  three  of  them  are  given, 
the  other  two  may  be  found. 

560.  To  find  the  Last  Term,,  the  First  Term,  the  Number 
of  Terms,  and  the  Common  Difference  being  given. 

i.  Required  the  last  term  of  the  ascending  series  having 
7  terms,  its  first  term  being  3,  and  its  common  difference  2. 

ANALYSIS. — From  the  definition,  each  succeeding  term  is  found 
by  adding  the  common  difference  to  the  preceding.  The  series  is : 

3,    3  +  2,    3 +  (2 +  2),    3 +  (2 +  2  +  2),    3  + (2  +  2  +  2 +  2),  etc.     Or 
3>    3  +  2,    3 +  (2x2),    3 +  (2x3),         3 +  (2x4),  etc.    That  is, 

561.  The  last  term  is  equal  to  the  first  term,  increased  by 
the  product  of  the  common  difference  into  the  number  of  terms 
less  i.  Hence,  the 

RULE. — Multiply  the  common  difference  by  the  number 
of  terms  less  i,  and  add  the  product  to  the  first  term. 

556.  What  is  an  arithmetical  progression  ?  Note.  The  first  and  last  terms 
called?  The  intervening ?  5^7.  The  common  difference?  5^8.  An  ascending 
pcries?  Descending?  560.  How  And  the  last  term,  when  the  first  term,  th$ 
number  of  terms,  and  the  common  difference  are  given  ? 


ARITHMETICAL     PROGRESSION.  351 

ES. — i.  In  a  descending  series  the  product  must  be  subtracted 
from  the  first  term  ;  for,  in  this  case  each  succeeding  term  is  found 
by  subtracting  the  common  difference  from  the  preceding  terms. 

2.  Any  term  in  a  series  may  be  found  by  the  preceding  rule. 
For,  the  series  may  be  supposed  to  stop  at  any  term,  and  that  may 
be  considered,  for  the  time,  as  the  last. 

3.  If  the  last  term  is  given,  and  the  first  term  required,  invert  the 
order  of  the  terms,  and  proceed  as  abate. 

2.  If  the  first  term  of  an  ascending  series  is  5,  the 
common  difference  3,  and  the  number  of  terms  1 1,  what 
is  the  last  term?  Ans.  35. 

3.  The  first  term  of  a  descending  series  is  35,  the  com- 
mon difference  3,  and  the  number  of  terms  10:  what  is 
the  last  ? 

4.  The  last  term  of  an  ascending  series  is  77,  the  number 
of  terms  1 9,  and  the  common  difference  3 :  what  is  the 
first  term  ?  Ans.  23. 

5.  What  is  the  amount  of  $150,  at  1%  simple  interest, 
for  20  years  ? 

562.    To  find  the  Number  of  Terms,  the  Extremes,  and 
the  Common  Difference  being  given. 

6.  The  extremes  of  an  arithmetical  series  are  4  and  37, 
and  the  common  difference  3 :  what  is  the  number  of  terms  ? 

ANALYSIS. — The  last  term  of  a  series  is  equal  to  the  first  term 
increased  or  diminished  by  the  product  of  the  common  difference 
into  the  number  of  terms  less  i.  (Art.  561.)  Therefore  37—4,  or 
33,  is  the  product  of  the  common  difference  3,  into  the  number  of 
terms  less  i.  Consequently  33-f-  3  or  1 1,  must  be  the  number  of  terms 
less  i ;  and  n  +  i,or  12, is  the  answer  required.  (Art.  93.)  Hence,  the 

RULE. — Divide  the  difference  of  the  extremes  by  the  com- 
mon difference,  and  add  i  to  the  quotient. 

7.  The  youngest  child  of  a  family  is  i  year,  the  oldest, 
21,  and  the  common  difference  of  their  ages  2  y. :  how 
many  children  in  the  family  ? 


562.  How  find  the  number  of  terms,  when  the  extremes  and  common  different 
(ire  given  ? 


352  ARITHMETICAL     PROGRESSION. 

563.  To  find  the  Common  Difference,  the  Extremes  and 

the  Number  of  Terms  being  given. 

8.  The  extremes  of  a  series  are  3  and  21,  and  the  number 
of  terms  is  10:  what  is  the  common  difference? 

ANALYSIS. — The  difference  of  the  extremes  21  — 3=18,  is  the  pro- 
duct of  the  number  of  terms  less  i  into  the  common  difference,  and 
10—  i,  or  g,  is  the  number  of  terms  less  i ;  therefore  18-^-9,  or  2,  is  the 
common  difference  required.  (Art.  93.)  Hence,  the 

EULE. — Divide  the  difference  of  the  extremes  by  the 
number  of  terms  less  i. 

9.  The  ages  of  7  sons  form  an  arithmetical  series,  the 
youngest  being  2,  and  the  eldest  20  years:  what  is  the 
difference  of  their  ages  ? 

564.  To  find  the  Sum   of  all  the  Terms,  the   Extremes  and 

the  Number  of  Terms  being  given. 

10.  Eequired  the  sum  of  the  series  having  7  terms,  the 
extremes  being  3  and  15. 

ANALYSIS. — (i.)  The  series  is  3,     5,     7,     9,   n,    13,   15. 

(2.)  Inverting  the  same,     15,    13,    n,      9,      7.      5,      3. 

(3.)  Adding  (i.)  and  (2.),      184-18  +  i8~+ 18  +  18  +  18  + 18= twice  the  sum. 

(4.)  Dividing  (3.)  by  2,        9  +  9  +  9  +  9  +  9  +  9  +  9=63,  the  sura. 

By  inspecting  these  series,  we  discover  that  half  the  sum  of  the 
extremes  is  equal  to  the  average  value  of  the  terms.  Hence,  the 

RULE. — Multiply  half  the  sum  of  the  extremes  by  the 
number  of  terms. 

REMARK. — From  the  preceding  illustration  we  also  see  that, 
The  sum  of  the  extremes  is  equal  to  the  sum  of  any  two  terms 

equidistant  from  them ;  or,  to  twice  the  sum  of  the  middle  tern; ,  if 

the  number  of  terms  be  odd. 

11.  How  many  strokes  does  a  common  clock  strike  in 
12  hours? 

563.  How  find  the  common  difference  when  the  extremes  and  number  of  terms 
are  jriven  ?  564.  How  find  the  sum  of  all  the  terms,  when  the  extremes  &nd 
number  of  terms  are  given  ? 


GEOMETRICAL    PROGRESSION. 

565.  A  Geometrical  Progression  is  a  series  of 
numbers  which  increase  or  decrease  by  a  common  ratio. 

NOTE. — The  series  is  called  Ascending  or  Descending,  according 
as  the  terms  increase  or  decrease.  (Art.  558.) 

566.  IQ  Geometrical  Progression   there  are  also  five  parts  or 
elements  to  be  considered,  viz. :   the  first  term,  the  last  term,  the 
number  of  terms,  the  ratio,  and  the  sum  of  all  the  terms. 

567.    To  find  the  Last   Term,  the   First  Term,  the   Ratio, 
and  the  Number  of  Terms  being  given. 

1.  Required  the  last  term  of  an  ascending  series  having 
6  terms,  the  first  term  being  3,  and  the  ratio  2. 

ANALYSIS. — From  the  definition,  the  series  is 
3>    3X2,    3  x  (2  x  2),    3  x  (2  x  2  x  2),      3  x  (2  x  2  x  2  x  2),  etc.     Or 
3,    3x2,    3  x  22,  3  x  23,  3  x  24,  etc.     That  is, 

Each  successive  term  is  equal  to  the  first  term  multiplied  by  the 
ratio  raised  to  a  power  whose  index  is  one  less  than  the  number 
of  the  term.  Hence,  the 

RULE. — Multiply  the  first  term  by  that  power  of  the  ratio 
whose  index  is  i  less  than  the  number  of  terms. 

NOTES. — i.  Any  term  in  a  series  may  be  found  by  the  preceding  rule. 
For,  the  series  may  be  supposed  to  stop  at  that  term. 

2.  If  the  last  term  is  given  and  the  first  required,  invert  the  order 
of  the  terms,  and  proceed  as  above. 

3.  The  preceding  rule  is  applicable  to  Compound  Interest;  the 
principal  being  the  first  term  of  the  series ;  the  amount  of  $i  for  i 
year  the  ratio ;  and  the  number  of  years  plus  i,  the  number  of  terms. 

2.  A  father  promised  his  son  2  cts.  for  the  first  example 
he  solved,  4  cts.  for  the  second,  8  cts.  for  the  third,  etc. : 
what  would  the  son  receive  for  the  tenth  example  ? 

3.  What  is  the  amount  of  $1500  for  5  years,  at  6%  com- 
pound interest  ?     Of  $2000  for  6  years,  at  7%  ? 

565.  WTiat  is  a  geometrical  progression  ?  567.  How  find  the  last  term,  when 
the  flr«t  term,  the  ratio,  and  number  of  terms  are  given  ? 


354  GEOMETKICAL     PEOGEESSION". 

568.  To  find  the  Sum  of  all  the  Terms,  the  Extremes 

and  Ratio  being  given. 

4.  Required  the  sum  of  the  series  whose  first  and  last 
terms  are  2  and  162,  and  the  ratio  3. 

ANALYSIS. — Since  each  succeeding  term  is  found  by  multiplying 
the  preceding  terra  by  tlie  ratio,  tlie  series  is  2,  6,  18,  54,  162. 

(i.)  The  sum  of  the  series,   —21-6  +  18  +  54+162. 

(2.)  Multiplying  by  3,  =      6  +  18  +  54+162  +  486. 

(3.)  Subt.  (i.)  from  (2.),  486—2=484,  or  twice  the  sum. 

Therefore  484-4-2=242,  the  sum  required.  But  486,  the  last  term 
of  the  second  series,  is  the  product  of  162  (the  last  term  of  the  given 
series)  into  the  ratio  3  ;  the  difference  between  this  product  and  the 
first  term  is  486—2  or  484,  and  the  divisor  2  is  the  ratio  3—1. 
Hence,  the 

KULE. — Multiply  the  last  term  ~by  the  ratio,  and  divide 
the  difference  between  this  product  and  the  first  term  by  the 
ratio  less  i. 

5.  The  first  term  is  4,  the  ratio  3,  and  the  last  term  972  : 
what  is  the  sum  of  the  terms  ? 

6.  What  sum  can  be  paid  by  1 2  instalments ;  the  first 
being  $i,  the  second  $2,  etc.,  in  a  geometrical  series  ? 

569.  To  find  the  Sum  of  a  Descending  Infinite  Series,  the 

First  Term  and  Ratio  being  given. 

REMARK. — In  a  descending  infinite  series  the  last  term  being 
infinitely  small,  is  regarded  as  o.  Hence,  the 

RULE. — Divide  the  first  term  by  the  difference  between 
the  ratio  and  i,  and  the  quotient  will  be  the  sum  required. 

7.  What  is  the-  sum  of  the  series  f,  f,  £,  -fa,  continued 
to  infinity,  the  ratio  oeing  £?  Ans.  i}. 

NOTE. — The  preceding  problems  in  Arithmetical  and  Geometrical 
Progressions  embrace  their  ordinary  applications.  The  others  involve 
principles  with  which  the  pupil  is  not  supposed  to  be  acquainted. 

5<ss.  How  find  the  sum  of  the  terms,  when  the  extremes  and  ratio  are  given  ? 
.509.  How  find  the  sum  of  an  infinite  descending  series,  when  the  first  term  and 
ratio  are  given  ? 


MENSURATION. 

570.  Mensuration  is  the  measurement  of  magni- 
tude. 

571.  Magnitude  is  that  which  has  one  or  more  of 
the  three  dimensions,  length,  breadth,  or  thickness;   as, 
lines,  surfaces,  and  solids. 

A  Line  is  that  which  has  length  without  breadth. 
A   Surface  is   that  which  has   length  and  breadth, 
without  thickness. 

A  Solid  is  that  which  has  length,  breadth,  and.  thickness. 

572.  A  Quadrilateral  Figure  is  one  which  has 
four  sides  and/owr  angles. 

NOTES. — i.  If  all  its  sides  are  straight  lines  it  is  rectilinear. 

2.  If  all  its  angles  are  right  angles  it  is  rectangular. 

3.  Figures  which  have  more  than/ww  sides  are  called  Polygons. 

573.  Quadrilateral  figures  are  commonly  divided  into  the  rect- 
angle, square,  parallelogram,  rhombus,  rhomboid,  trapezium,  and 
trapezoid. 

]@p~  For  the  definition  of  rectangular  figures,  the  square,  etc.,  se« 
Arts.  240,  241. 

574.  -A-    Rhombus    is    a    quadrilateral 
which   has  all  its  sides  equal,  and  its  angles 
oblique. 

575.  A  Rhomboid  is  a  quadrilateral 
in  which  the  opposite  sides  only  are  equal,  and 
all  its  angles  are  oblique. 

576.  The  altitude  of  a  quadrilateral  figure 
having  two  parallel  sides  is  the  perpendicular 
distance  between  these  sides ;  as,  A,  L. 

577.  A  Trapezium  is  a  quadrilateral  which  has  only  two  of 
its  sides  parallel.*     (See  next  Fig.) 


570.  What  is  Mensuration  ?    571.  Magnitude  ?  Line  ?   Surface  ?   Solid  ?  574.  A 
•hombus  J     575.  Rhomboid  ?    576.  The  altitude  ?    577.  A  trapezium  ? 
*  Legendre,  Dr.  Brewster,  Young,  Du  Morgan,  etc. 


356  MENSURATION. 

578.  A  Diagonal  is  a  straight  line  joining  the 
vertices  of  two  angles,  not  adjacent  to  each  other'; 
as,  A  D. 

579.  The  common  measuring  unit  of  surfaces  is 
a  square,  whose  side  is  a  linear  unit   of  the  same 
name.     (Thomson's  Geometry,  IV.     Sch.) 

Jg^~  To  find  the  area  of  rectangular  surfaces,  see  Art.  280. 
The  rules  or  formulas  of  mensuration  are  derived  from  Geometry 
to  which  their  demonstration  properly  belongs. 

580.  To  find  the  Area  of  an  Oblique  Parallelogram,  Rhombus, 
or  Rhomboid,  the  Length  and  Altitude  being  given. 

Multiply  the  length  by  the  altitude. 

NOTE. — If  the  area  and  altitude,  or  one  side  are  given,  the  other 
factor  is  found  by  dividing  the  area  by  the  given  factor.     (Art.  244.* 

1.  What  is  the  area  of  a  rhombus,  its  length   being 
60  rods,  and  its  altitude  53  rods?  Ans.  3i8osq.  r. 

2.  A  rhomboidal  garden  is  75  yds.  long,  the  perpendicular 
distance  between  its  sides  48  yds. :  what  is  its  area  ? 

3.  The  area  of  a  square  field  whose  side  is  120  rods  ? 

4.  If  one  side  of  a  rectangular  grove  containing  80  acres 
is  1 60  rods,  what  is  the  length  of  the  other  side  ? 

581.  To  find  the  Area  of  a  Trapezium,  the  Altitude  and 

Parallel  Sides  being  given. 

Multiply  half  the  sum  of  the  parallel  sides  by  the  altitvde. 

5.  The  altitude  of  a  trapezium  is  n  ft-,  and  its  parallel 
sides  are  16  and  27  ft.:  what  is  its  area?    Ans.  236.5  sq.ft 

582.  To   find   the    Area    of  a    Triangle,   the    Base    and 

Altitude  being  given. 

Mutely  the  base  by  half  the  altitude.     (Art.  539.) 

NOTE. — Dividing  the  area  of  a  triangle  by  the  altitude  gives  the 
base.     Dividing  the  area  by  half  the  base  gives  the  altitude. 


578.  A  diagonal V  579.  The  common  measuring  null  of  surface??  580.  How 
find  the  area  of  an  oblique  parallelogram  or  rhombus?  583.  How  find  the  are* 
of  a  trapezium  ?  582.  Of  a  triangle  ? 


MENSURATION.  35? 

6.  What  is  the  area  of  a  triangle  whose  base  is  37  ft., 
and  its  altitude  19  ft.  ?  Ans.  351.5  sq.  ft. 

7.  Sold  a  triangular  garden  whose  base  is  50  yds.,  and 
altitude  40  yds.,  at  $2.75  a  sq.  rod:  what  did  it  come  to? 

583.  To  find  the  Circumference  of  a  Circle,  the  Diameter 

being  given. 

Multiply  the  diameter  by  3.14159. 

E2T  For  definition  of  the  circle  and  its  parts,  see  Art.  257. 

8.  The   diameter  of  a  cistern  is   1 2  ft. :   what  is  its 
circumference?  Ans.  37.69908  ft. 

9.  The  diameter  of  a  circular  pond  is  65  rods :  what  is 
its  circumference  ? 

584.  To   find    the  Diameter   of  a   Circle,  the   Circum- 

ference being  given. 

Divide  the  circumference  by  3.14159. 

10.  What  is  the  diameter  of  a  circle  whose  circum- 
ference is  150  ft.  ? 

1 1.  The  diameter  of  a  circle  100  rods  in  circumference  ? 

585.  To  find    the   Area  of  a   Circle,  the    Diameter   and 

Circumference  being  given. 

Multiply  half  the  circumference  by  half  the  diameter. 

12.  Eequired  the  area  of  a  circle  whose  diameter  is  75  ft. 
xi3.  What  is  the  area  of  a  circle  200  r.  in  circumference  ? 

586.  The  common  measuring  unit  of  solids  is  a  cube, 
whose  sides  are  squares  of  the  same  name.  The  sides  of 
a  cubic  inch  are  square  inches,  etc. 

jgjp  For  definition  of  rectangular  solids,  and  the  method  of  find- 
ing their  contents,  see  Arts.  246,  247,  and  284. 

583.  How  find  the  circumference  of  a  circle  when  the  rliamcter  is  given* 
^84.  How  find  the  diameter  of  a  circle,  when  the  circumference  is  given? 
585.  How  find  the  area  of  a  circle  ?  586.  What  is  the  men* uring  unit  of  solids  ? 


358 


MEN  SITUATION. 


587.  A  Pyramid    is   a    solid 
whose  base  is  a  triangle,  square,  or 
polygon,  and  whose  sides  terminate 
in  a  point. 

NOTE. — This  point  is  called  the  vertex  of 
the  pyramid,  and  the  sides  which  meet  in 
it  are  triangles. 

588.  A  Cone  is  a  solid  which 
has  a  circle  for  its  base,  and  termi- 
nates in  a  point  called  the  vertex. 

589.  A  Frustum  is  the  part 
which  is  left  of  a  pyramid  or  cone, 
after  the  top  is  cut  off  by  a  plane 
parallel  to  the  base ;  as,  a,  It,  c,  d,  c. 


590.  To  find  the  Contents  of  a  Pyramid  or  Cone,  the 

Base  and  Altitude  being  given. 

Multiply  the  area  of  the  base  by  %  of  the  altitude. 

XOTH. — The  contents  of  a  frustum  of  a  pyramid  or  cone  are  found 
by  adding  the  areas  of  the  two  ends  to  the  square  root  of  the  product  of 
those  areas,  and  multiplying  the  sum  by  ^  of  the  altitude. 

14.  What  are  the  contents  of  a  pyramid  whose  base  is 
22  ft.  square,  and  its  altitude  30  ft.  Ans.  4840  cu.  ft 

15.  Of  a  cone  45  ft.  high,  whose  base  is  18  ft.  diameter? 

1 6.  The  altitude  of  a  frustum  of  a  pyramid  is  32  ft, 
the  ends  are  5  ft.  and  3  ft  square :  what  is  its  solidity  ? 

591.  A  Cylinder  is  a  roller-shaped  solid  of  uniform 
diameter,  whose  ends  are  equal  and  parallel  circles. 

592.  To  find  the  Contents  of  a  Cylinder,  the  Area  of  the 
Base  and  the  Length  being  given. 

Multiply  the  area  of  one  end  by  the  length. 

17.  "What  is  the  solidity  of  a  cylinder  20  ft.  long  and 
4  ft  in  diameter?  Ans.  251.3272  cu.  ft. 

5*7.  What  is  a  pyramid?    588.  A  cone!    590.  How  find  the  contents  of  each? 


MENSUBATION.  359 

593.    To  find  the   Convex  Surface  of  a   Cylinder,  the 

Circumference  and  length  being  given. 

Multiply  the  circumference  by  the  length. 

1 8.  Eequired   the  convex  surface  of  a  cylindrical  log 
whose  circumference  is  18  ft.,  and  length  42  ft.? 

594.  A  Sphere  or  Globe  is  a  solid  terminated  by  a 
curve  surface,  every  part  of  which  is  equally  distant  from 
a  point  within,  called  the  center. 

595.  To  find  the  Surface  of  a  Sphere,  the  Circumference 

and  Diameter  being  given. 

Multiply  the  circumference  by  the  diameter. 

19.  Eequired  the  surface  of  a  15  inch  globe.  Ans.  4.91  sq.ft. 

20.  Required  the   surface  of  the   moon,  its  diameter 
being  2162  miles. 

596.  To  find  the  Solidity  of  a  Sphere,  the  Surface  and 

Diameter  being  given. 

Multiply  the  surface  by  %  of  the  diameter. 

21.  "What  is  the  solidity  of  a  10  inch  globe?  A.  523.60^  in. 

22.  What  is  the  solidity  of  the  earth,  its  surface  being 
197663000  sq.  miles,  and  its  mean  diameter  7912  miles? 

597.  To  find  the  Contents  of  a  Cask,  its  length  and  head 

diameter  being  given. 

Multiply  the  square  of  the  mean  diameter  by  the  length, 
and  this  product  by  .0034.  The  result  is  wine  gallons. 

NOTES. — i.  The  dimensions  must  be  expressed  in  inches. 

2.  If  the  staves  are  much  curved,  for  the  mean  diameter  add  to  the 
head  diameter  .7  of  the  difference  of  the  head  and  bung  diameters ; 
if  little  curved,  add  .5  of  this  difference  ;  if  a  medium  curve,  add  .65. 

23.  Required  the  contents  of  a  cask  but  little  curved, 
whose  length  is  48  in.,  its  bung  diameter  36  in.,  and  its 
head  diameter  34  inches.  Ans.  199.92  gal. 


591.  What  is  a  cylinder ?    592.  How  find  its  contents?    594.  What  is  a  sphere 
or  globe  ?    593.  How  find  its  surface  ?  596.  Its  contents  T    597.  Contents  of  a  cask  T 


360  MISCELLANEOUS     EXAMPLES. 


MISCELLANEOUS     EXAMPLES. 

1.  A  square  piano  costs  $650,  which  is  3-fifths  the  price 
of  a  grand  piano :  what  is  the  price  of  the  latter  ? 

2.  A  man  sold  his  watch  for  $75,  which  was  5 -eighths 
of  its  cost :  what  was  lost  by  the  transaction  ? 

3.  Two  candidates  received  2126  votes,  and  the  victor 
had  742  majority :  how  many  votes  had  each  ? 

4.  A  man  owning  3  lots  of  154,242  and  374  ft.  front 
respectively,  erected  houses  of  equal  width,  and  of  the 
greatest  possible  number  of  feet :  what  was  the  width  ? 

5.  Three  ships  start  from  New  York  at  the  same  time 
to  go  to  the  West  Indies ;  one  can  make  a  trip  in  10  days, 
another  in  12  days,  and  the  other  in  16  days:  how  soon 
will  they  all  meet  in  New  York  ? 

6.  Two  men  start  from  the  same  point  and  travel  in 
opposite  directions — one  goes  33^  m.  in  7  h. ;  the  other 
27-^  m.  in  5  h. :  how  far  apart  will  they  be  in  14  h.  ? 

7.  Divide  $200  among  A,  B  and  C,  giving  B  twice  as 
much  as  A,  and  C  3 1  times  as  much  as  B. 

8.  How  many  bushels  of  oats  are  required  to   sow 
35f  acres,  allowing  i  i-J  bushels  to  5  acres  ? 

9.  Bought  4-|  bbls.  of  apples  at  $3$,  and  paid  in  wood 
at  $3!:  a  cord :  how  many  cords  did  it  take  ? 

10.  A  mason  worked  u|  days,  and  spending  f  of  his 
earnings,  had  $20  left:  what  were  his  daily  wages? 

11.  A  fruit  dealer  bought  5250  oranges  at  $31.25  per  M., 
and  retailed  them  at  4  cents  eacli :  what  did  he  make  or  lose  ? 

12.  Cincinnati  is  7°  50'  4"  west  of  Baltimore:   when 
noon  at  the  former  place,  what  time  is  it  at  the  latter  ? 

13.  A  grocer  bought  1000  doz.  eggs  at  12  cts.,  and  sold 
them  at  the  rate  of  20  for  25  cts. :  what  was  his  profit  ? 

14.  Bangor,  Me.,  is  21°  13'  east  of  New  Orleans:  when 
9  A.  M.  at  Bangor,  what  is  the  hour  at  New  Orleans  ? 

15.  What  is  the  cost  of  a  stock  of  i  ^boards  15  ft.  long 
and  i  o  in.  wide,  at  1 6  cts.  a  foot  ? 


MISCELLANEOUS     EXAMPLES.  361 

1 6.  A   farmer  being  asked   how  many  cows  he  hud, 
replied  that  he  and  his  neighbor  had  27  ;  and  that  f  of  his 
number  equaled  -f-y  of  his  neighbors :  how  many  had  each  ? 

17.  How  many  sheets  of  tin  14  by  20  in.  are  required  to 
cover  a  roof,  each  side  of  which  is  25  ft.  long  and  2 1  ft.  wide  ? 

1 8.  A  and  B  counting  their  money,  found  they  had 
$100 ;  and  that  £  of  A's  plus  $6  equaled  f  of  B's:  how 
much  had  each  ? 

19.  How  many  pickets  4  in.  wide,  placed  3  in.  apart,  are 
required  to  fence  a  garden  21  rods  long  and  14  rods  wide  ? 

20.  Bought  a  quantity  of  tea  for  $768,  and  sold  it  for 
$883.20 :  what  per  cent  was  the  profit  ? 

21.  What  must  be  the  length  of  a  farm  which  is  80  rods 
wide  to  contain  75  acres  ? 

22.  What  must  be  the  height  of  a  pile  of  wood  36  ft.  long 
and  12  feet  wido  to  contain  27  cords  ? 

23.  Sold  60  bales  of  cotton,  averaging  425  lb.,  at  22^  cts., 
on  9  m.,  at  1%  int. :  what  shall  I  receive  for  the  cotton  ? 

24.  When  it  is  noon  at  San  Francisco,  it  is  3  h.  i  m. 
39  sec.  past  noon  at  Washington :  what  is  the  difference 
in  the  longitude  ? 

25.  A  man  bought  a  city  lot  104  by  31  \  ft.,  at  the  rate 
of  $22^  for  9  sq.  ft. :  what  did  the  lot  cost  him  ? 

26.  If  ^J  of  a  ton  of  hay  cost  £3  £,  what  will  |-£  ton  cost  ? 

27.  What  sum  must  be  insured  on  a  vessel  worth  $16500, 
to  recover  its  value  if  wrecked,  and  the  premium  at  2%? 

28.  How  many  persons  can  stand  in  a  park  20  rods  long 
and  8  rods  wide ;  allowing  each  to  occupy  3  sq.  ft.  ? 

29.  A  builder  erected  4  houses,  at  the  cost  of  $4284-} 
each,  and  sold  them  so  as  to  make  16%  by  the  operation: 
what  did  he  get  for  all  ? 

30.  A  man  planted  a  vineyard  containing  16  acres,  the 
vines  being  8  ft.  apart :  what  did  it  cost  him,  allowing  he 
paid  6 \  cts.  for  each  vine  ? 

31.  If  a  perpendicular  pole  10  ft.  high  casts  a  shadow 
of  7  ft.,  what  is  the  height  of  a  tree  whose  shadow  is  54  ft  ? 


362  MISCELLANEOUS     EXAMPLES. 

32.  A  miller  sold  a  cargc  of  flour  at  20%  profit,  by  which 
he  made  $2500:  what  did  he  pay  for  the  flour? 

33.  Paid  $1.25  each  for  geographies:  at  what  must  I 
mark  them  to  abate  6%,  and  yet  make  20%  ? 

34.  Sold  goods  amounting  to  $1500,  |  on  4  m,  the 
other  on  8  m. ;  and  got  the  note  discounted  at  •]%:  what 
were  the  net  proceeds  ? 

35.  A  line  drawn  from  the  top  of  a  pole  36  ft.  high  to 
the  opposite  side  of  a  river,  is  60  ft  long:  what  is  the 
width  of  the  river  ? 

36.  A  school-room  is  48  ft.  long,  36  ft.  wide,  and  1 1  ft. 
high :  what  is  the  length  of  a  line  drawn  from  one  corner 
of  the  floor  to  the  opposite  diagonal  corner  of  the  ceiling  ? 

37.  The  debt  of  a  certain  city  is  $212624.70:  allowing 
(>%  for  .collection,  what  amount  must  be  raised  to  cover 
the  debt  and  commission  ? 

38.  A  publisher  sells  a  book  for  62^  cents,  and  makes 
20%'.  what  per  cent  would  he  make  if  he  sold  it  at  75  cents? 

39.  What  will  a  bill  of  exchange  for  £534,  los.  cost  in 
dollars  and  cents,  at  $4.87!  to  the  £  sterling  ? 

40.  Three  men  took  a  prize  worth  $27000,  and  divided 
it  in  the  ratio  of  2,  3,  and  5  :  what  was  the  share  of  each  ? 

41.  An  agent  charges  5^  for  selling  goods,  and  receives 
$135.50  commission  :  what  are  the  net  proceeds? 

'42.  A,  B,  and  C  agreed  to  harvest  a  field  of  corn  for 
$230;  A  furnished  5  men  4  days,  B  6  men  5  days,  and 
C  7  men  6  days:  what  did  each  contractor  receive? 

43.  A  and  B  have  the  same  salary ;  A  saves  £  of  his, 
but  B  spending  $40  a  year  more  than  A,  in  5  years  was 
$50  in  debt:  what  was  their  income,  and  what  did  each 
spend  a  year  ? 

44.  If  a  pipe  6  inches  in  diameter  drain  a  reservoir  in 
80  hrs.,  in  what  time  would  one  2  ft  in  diameter  drain  it  ? 

45.  A  young  man  starting  in  life  without  money,  saved 
$i  the  first  year,  $3  the  second,  $9  the  third,  and  so  on, 
for  12  vears:  how  much  was  he  then  worth? 


ANSWERS. 


ADDITION. 


Ex.    ANS. 

Ex.    ANS. 

Ex.    Aus. 

Ex.    ANS. 

J.  age  <$*>. 

14.  2616263 

33.  13800 

Page  29. 

2.  13839 
3.  l82<0 

15-  9539381 

1  6.  $4668 

Page  28. 

$33658,  all- 

O           -J 

4.  20000 

34-  159755 

.   o  o  _.  ^ 

54.  156  str. 

5-  20438 
6.  212269 

Page  27. 
17.  1376  yds. 

35.  848756 

36.  182404 

37.  1039708 

55.  7213  bu. 
56-  366  d. 

Page  2G. 

1  8.  $6332 
19.  1695  Ibs. 

38.  11485 
39.  9929 

58.  50529 

CQ.  -2674,  A. 

i.  2806 

20.  2668  g. 

40.  13720 

0  y'  o  I  ^  **• 

60.  $4.37.4.4, 

2.  $1941 

21.  10438 

41-  233331 

*-'•  virO  /  T1  r 

3.  25285  Ibs. 

4-  14756  yds. 

22.  8636 

23.  10672 

42.  1328464 
43.  8237027 

62.  $376.02 
63.  $476.19 

5-  98937  r. 

24.  3874 

44.  25148 

6.  2051834  ft 
7.  2460  A. 

25.  15246 

26.  100980 

45.  limiio 
46.  22226420 

65.  $475.89 

8-  23459 

27.  1207053 

47.  $1460 

Page  3O. 

9.  185462 

28.  $9193 

48.  1925  y. 

66.  $1704.28 

10.  76876 

29.  3998  bn. 

49.  $8190 

67.  $16988.71 

"•  33367 

30.  $107601 

50.  6987  Ibs. 

68.  $16580.34 

12.  179589 

31.  $38058 

51-  93  yrs- 

69.  $179403.71 

13.  273070 

32.  2844     '52.  $22338   170.  $157011.73 

SUBTRACTION. 

Page  35. 

4-  3779  7- 

15.  309617  T. 

Page  37. 

2.  346 

5-  J7i9  A. 

16.  209354  A. 

26.  $279979 

3-  H7 

6.  11574 

J7-  34943 

27.  iiiiiu 

4-  3IO(5 

7-  22359 

1  8.  1235993 

28.  6333333334 

5-  26°3 

8.  27179 

r9-  3633805 

29.  mil  1  1  112 

6.  509 

9.  267642 

20-  3323°°76 

30.  289753017- 

10.  235009 

21.  349629696 

746. 

Page  3d. 

II.  5009009    1  22.  $18990     |3I.  270305844* 

'•  53637 

12.  5542809    |  23.  $1915             28516. 

2.  3°5  r- 

13.  2738729 

24.  $415026    J32.  226637999- 

3.  67  Ibs. 

14.  51989  Ibs. 

25.  $2OOOO5 

876130. 

364 


AN8WEBS. 


Ex.    ANS. 

Ex.    ANS. 

Ex.    ANS.    i  Ex.    ANS. 

33.  1990005 
34.  995500 
35.  64564 

36.  999001000 

39.  2235  A. 

40.  26530000 

41.  1732  yrs. 
42.  84  yrs. 
43.  1642  y. 

44.  413000000 
45.  $3115027 
46.  8253204 

QUESTIONS     FOR     REVIEW. 


Page  38. 

7-  3771 

Page  39. 

23.  2198 

8-  803559 

24.  $4730 

I.  $1200 

9.  $14292 

16.  108888 

25.  $522. 

2.  $1107 

10.  $963 

17.  126950 

?6.  $1802 

3.  2365  sold, 

ii.  $12523 

18.  38022 

27.  $13970  B's 

1195  left. 

12.  1700 

19.  43898 

$27440  C's 

4.  $4331 

13-  4434 

20.  33885 

28.  5673 

5-  447i 

14.  1  1  21  22 

21.  762 

29-  $737 

6.  1279 

15.  127680 

22.  l68o 

30.  $18775 

MULTIPLICATION. 


Page  46. 

2.    10224 

3-  1970S 

4.  64896 

5.  7.61824 

Page  48. 


16.  2923420500 

17.  1572150300 

18.  450029050401 


19.  1924105179680 

20.  86625  Iks. 

21.  $5535° 

22.  22360  bu. 

23.  $222984 

24.  78475  m. 

25.  1351680  ft 

26.  §48645 

27.  $33626 

28.  $2367500 

29-  54075  yds, 

30.  $1786005 

31.  $242284 

32.  432  m. 


Page  47. 

2-  242735 
3.  1230710 

4-  3627525 
5.  20136672 

Page  5O. 

3.  65100 

4.  290496 

5-  509733 
6.  3263112 


Page  48. 

6.  332671482 

7.  941756556 

8.  1524183620 

9.  2751320848 


10.  131247355 

11.  351410400 

12.  532125480 

13.  1651148750 


7.  14515201.111. 

8.  $295008 

9.  $197500 


Page  SI. 

21.  I49IOOO 

22.  3328000 

23.  166092000 

24.  740000  Ibs. 

25.  $184000 

26.  2600000  cts. 

27.  $1050000 

28.  604800000  t.  i  46.  454842 

29.  $7350000    47.  1585873 


•  I5833l855° 

30.  40226351915148000 

31.  64003360044100000 

32.  22812553589100000 

Page  52. 

33.  12615335010000000 

34.  312159700930000000 

35.  263200000756000000 

36.  20776000000000 

37.  42918958404000 

38.  370475105000000 

39.  652303302651 

40.  400800840440000 
42.  17784 

43-  29484 

44-  45975 

45-  107872 


ANSWERS. 


365 


SHORT     DIVISION. 


Ex.    ANS. 

Ex.   ANS. 

Ex.    ANS. 

Ex.    ANS. 

Paye  59. 

13.  8243  h. 

22.  1280604^ 

34.  9065}  t. 

2.  218392 

14.  211  S. 

23.  IOOII62 

35.  52483-  bar. 

•3.  186782 

15.  8978f  r. 

24.  746367rr 

36.  343^  F- 

4.  I72258| 

16.  $10671 

25.  1200381  . 

37-  6469f  yds. 

5.  1496471 

17.  122  yds. 

26.  2346842 

38.  $9405 

6.  662107^ 

18.  14140^  Ibs 

27.  3562695^ 

39.  5812  sq.yd. 

7.  6865867 

19.  $8121 

28.  5848142! 

40.  2500  hrs. 

8.  923808 

29.  84472327 

41.  70440  b. 

9.  922969! 
10.  5762314 

Page  GO. 

30.  59363694^ 
31.64519169^ 

42.  $i453r 
43-  335  183  A. 

ii.  60663768-^ 

20.  1067102^ 

32.  3794|  w. 

44.  13920!  bar. 

12.  680O2  1  033-1^ 

2I-  933539 

33.  $6412 

45.  8090  cows 

LONG     DIVISION. 


Page  63. 

3-  2312^ 
4.  4091 
5-  2076^ 

/\      f\  I  3 

7- 
8. 

9-  9759if 

10.  1025277 

11.  50872!^ 


-5 


12. 

13- 

14.  1080887^ 

17-  24377 
18.  454  am. 

Page  64. 

20.  29796^ 


21. 

22. 

23- 
24. 

25- 
26. 
27. 
28. 
29. 
3°- 

32. 

33- 

34- 
35- 
36. 
37- 

38. 


I2294ffi 

450  sh. 
411^8.  ft. 

$94 


54<>H* 

$60  2 


cu.ffc. 


Page  65. 

$4000 


39- 

40. 

41. 

42. 

43- 

2. 

3- 
4- 

5- 
6. 

7- 


$379°7iM 

$8050^^ 

$275 


66. 

17- 

•ft ;  19  or. 
2  3  pounds. 
12  com. 

17 
23 
26 


Page  67. 

9.  Given 


10. 

ii-  7974H 


12. 


14.  66042  if|- 

15.  1502085^ 

Page  69. 

24-  2283^ 

25.  4o6if>£ 

26. 

27. 

28. 

29. 

3°- 


33.  229  horses 

34- 

35- 

36.  60  bales 


366 


ANSWERS. 


QUESTIONS     FOR     REVIEW. 


Page  6'.9. 

i.  146  J's; 
365  both 
2.  407  sheep 
3.  76  years 
4.  1779 

5-  3°93 
6.  83^4 

/.    4,5  JLUUQ 

8.  5118  bu. 

9.  8613 

10.  35 
ii.  200849 

Page  7O. 

12.  $631  gain 

15.  i22-r*y  bar. 
1  6.  $i537i 
17.  60  days 
18.  $156 
19-  1  05^  yds. 

20.    $2 

a.  T5J033: 
23.  252  sheep 
24.  41  J  bar. 

25-  HT& 
26.  2383/7 
27.  482  books 
28.  13  cows 

PROBLEMS     AND    FORMULAS. 

Page  73-6. 

2.  1157  votes 
4.  1147  votes 
6.  1  60  rods 

8.  31  miles 
10.  $22680 

12.    60 

14.  $2196,  ist. 
$3172,  2d. 

15.  2428,  ist 
3136  2d. 
16.  $104  ch. 
$146  w. 

17.  45,  A's. 
30,  B's. 
1  8.  §269,  A's. 
$231,  B's. 

ANALYSIS. 

Page  78. 

i.  15  cows 
2.  50  Ibs. 
3.  432  Ibs. 
4.  60  bu. 
5-  31  Iks. 

6.  378  bu. 
7.  $  1  80  cloth; 

$10 

8.  Si8 
9.  $315 
10.  1  8  pears 

Page  7.9. 

12.    31 

13-  5°4 
14.  ii  chil. 
1  6.  12  less, 
60  greater 

17.  Given 
18.  $112  B's. 
6361  A's. 
19.  67,  ist 
176,  2d. 

20.    2Il8 

FACTORING. 

Page  88. 

23.  2  and  3 
24.  2  and  3 

25.  2  and  2 
26.  None. 
27.  2,  2,  and  2 
28.  2,  3,  and  2 

29.  5  and  3 
30.  2  and  2 
31.  2  and  2 

32.  5 
33.  2  and  2 
34.  2  and  3 

35-  !>  2,  3,  5,  7,  n,  13,  17,  19,  23,  29,  31,  37,  41,  43,  47,  53, 

59,  61,  67,  71,  73,  79,  83,  89,  97. 

36.  From  100-200  are  101,  103,  107,  109,  113,  127,  131,  137 
i39>  *49>  IS1*  i57»  l63>  l67>  i73>  *79>  *8i,  191,  193,  197,  199- 


ANSWERS. 


307 


CANCELLATION. 


Ex.       ANS. 

Ex.       ANS. 

Ex.       ANS. 

Ex.       ANS. 

Ex.       ANS. 

Page  9O. 

8.  28 

13.    90 

18.  36* 

23.  i  So  ch. 

4.  if 

9.   18 

14."   78  J 

19-  378 

24.  315  bu. 

5-  8 

10.   15 

15.   84 

20.  30  bar. 

25.  28  yrs. 

6.  3 

II.    12$ 

1  6.   84 

21.  sfffc. 

26.  81  1. 

7-  9 

12.     10 

I7.   318^ 

22.  46^  bgsj 

COMMON     DIVISORS. 


Page  91. 

8.  5  and  3     _            .11.6 
i  °     Pane  94. 

9.  2  and  4                         12.  4 

1  8.  one 
19-  37 

3-  3 

6.  21            13.  12 

2O.    2 

4-  4>  3>  6 
5-  6 

Page  93.        7-  *5             M-  2 

8.     12                   15.    12 

21.    2040 

22.  1  8  yd. 

6.  7 

2.    12                       9.    3                     1  6.    2 

23.  21  each 

7.  10 

3.  24               10.  25             17.   192 

24.  8  A. 

MULTIPLES. 

Page  98. 

6.  600           10.  288           14.  55440 

17.  60  cts. 

3-  48 

7.  480           n.   12852       15.  57600 

1  8.  720  ro's 

4.  84 

8.  330           12.  15120      16.  Jooo 

19.  60  Ibs. 

5.  720 

9.  240           13.  73440 

FRACTIONS. 

Page  1O5. 

15-  H 

29.  ^ 

13.  5^ 

27.  $29ffff 

2.    | 

16.  ii 

14.  46-^5- 

28.  3516  yr. 

3-  f 

17.  « 

Page  1O5. 

15.   iSflf 

4.1 

18.  | 

2-  37 

1  6.  gojfr 

Page  1O6. 

5-  f 

19.  T^y 

3.  1  6£ 

17.  22-,-^ 

2.   IfA 

6.  j 

20.    || 

4.  19 

18.    2IOJ 

3-  V 

7-  ii 

21.    | 

5-  :5l 

19.  io7TyT 

4.  -^l1- 

8-  i 

22.    |££ 

6.    12 

20.    38311 

s-^tt1 

9-1 

23-1 

7.  16 

21.    20| 

6.  2^^ 

10.  | 

24-1 

8.    I2| 

22.    2449^4 

7.  1W1 

ii.  Hi 

25-    1 

9-  36fi 

23.    lo||4&   ;     8.  -2^^ 

12.    f 

26.    3 

10.  45^ 

24.    29^1 

9.   -S^f^S. 

T3-  * 

•7-4 

11.  61 

25-  ^frlM 

10.     'R/g 

14.  £ 

28.  -^        12.  3^4 

26.  41-A^lb   ii.  J  '2-3-i 

ANSWERS. 


Ex.        ANS. 


12. 


3-267-2 
S3 


1 6. 

1  7-    I006T3T 

1 8.  ±2£Llbs. 


Ex.       ANS. 


9.  28 
10.  |f 

»•  3% 


Ex. 


ANS. 


-m 


Piojye  1O  7. 

3-  i 

4-  if 

5-  f 
6.  f 

7-  H 


T  f  3 

!5-  s 

1 6.  ?, 

17.  i 


20. 
21.4 

22.    |J  bll. 

23-  6s?  yds. 


I4' 


II    , 


T& 

Hf 

iff 

144 

^60^ 
650 

TOOCTO- 


T 


Ex. 
1 6. 


ANK. 


Ex.      ANS. 


1O9. 


9- 

I0. 
n. 

12. 
13- 


Page  <//<?. 

4-    7a5 

r       Jj 6 

6.  f 

7-  f 
8.  -f 

9-1 

10.  ff 

11.  -ff 

12.  f 

13-    lit 
14.    iff 


Page  111 

2-  if,  A 

3-  ff,  *f,  ** 

4-  iff> 

5-  -&5v> 
6- 

7. 

8. 


.  if 


*f,  Hff  H, 

Pttye  112. 


17. -Hi*,  *f  IMS 
18.  f!|,  ifi  |H 

=o.  if,  H  f» 

21-  ff,ff,ff    . 

22-  ff,    U 

23.  W,  ft 

, ,      Tns      QR        Q I 
2 4«    1irB»  T5"S>  7T¥ 

25*    1000,  TtfVo",  TO 00 


Page  113. 

3.  f  i  «,  ft 

4-  £>  i  i 

5-  -if,  tt,  If 

6-  f&, 

7-  if,  if,  -2A°- 

8-  «t,  Mi 

n       25        «        7 
y.    TTJ'  To",  T 


O      I  7  S 
">     20~ 


304      154 
26<f,  21T6 

1  *•    "JTT,  IjV,  1HJ, 

T  o     _SJ>         140 
I2-    ffif,     2TC> 

TlO,   2Tff 

13-  W,  ¥i  _ 

15. 45,'  w,  H 


17.  tHf,mt»mi 

TQ      32      53      4" 
10"     81),  TO,    89 


ANSWEBS. 


369 


ADDITION    OF    FRACTIONS. 


Ex.    ANS. 


Ex.     ANS. 


Ex.     ANS. 


Ex.     ANS. 


Ex.     ANS. 


Page  lid. 

8,  9.  Given 

10. 

ii. 


12. 
13- 

14. 

'5- 
1  6. 


17.  320  £ 

1  8.  75  if! 
19-  554-BtJ 


22.    I3TV 

23- 

24. 
25. 


26.  51  J  y. 

27.  $489| 

29- 

30. 


SUBTRACTION     OF     FRACTIONS. 


Page  118. 

10.  fj 

11.  2£ 

12.  IT7^ 
!4-    H 


16.  3tt* 


19. 

20.  32-!  Ibs. 

21.  87^  A. 


Page  119. 

27.    102-f^- 

34-  53 

28.    £ 

35-  I74H 

22.  Given 

29.  Given 

36.  iu-& 

23-  38i 

30.  i 

37.  40-!  gaJ 

24.  37i 

31*   TTZ 

38.  $29$ 

25-  43* 

32.  44 

40-  i-H- 

26'  67*        '  33-  5l 

41-  ifl 

MULTIPLICATION     OF     FRACTIONS. 


JFVeflre  /^J, 

3-  2j 

4-  7^r 

6-  3571 

7-  583i 

8.  36 

9.  277! 


u.  2566]- 

'12. 


13-  2.75 

14.  $6of 

15.  $231! 

16.  $1218^ 

17.  $8o4f 


1  8.  $1592^ 


20. 

21.    $S4l8 

Page  123. 

3-  24! 
4.  26^ 

5-  336 
6.  406 

7-  542$ 


10.  $6429!- 

12.    1407 


14.  1256^ 

15.  6i2j 

16.  1009! 

17.  21721-^ 

1 8.  70003! 

Page  124. 

3-1 

4-  i 

5-  6| 
6.  |4 

7-1 
8.  | 

10.  ii 


II. 


15-  3» 

1 6.  78^ 

1 7.  9  cts. 

1 8.  $36 

20.  703! 

2L  19531 

22.  15352^ 

24-  i 

25-  iV 
J'wfire  ^^5. 


4-  149^ 

5-  37  ii 

6.  600 

7.  3986ff 

8.  12377^5 
9-  I7317H 

n.  u|J 
12.  $5of 
13-  433 i -eta 

15.  750  tin, 
3000  cop. 

16.  $1918! 

X7-  44839^ 


370 


ANSWERS. 


DIVISION     OF     FRACTIONS. 


Ex.      ANS. 

Ex.      ANS. 

Ex.       Ass. 

Ex.      ANS. 

Ex.       ANS. 

Page  126. 

28.  68£f 

Page  ISO. 

O*   ^     o 

II.  -jVj 

*  T¥ 

30.  uSiVjr 

2.2|f 

4.  i7f^ 
5.  15  times 

12.  50^  bu. 

13.    245  I  T?0f. 

5-    <T7/ 

6-  *4y 

7-  "z2? 

3  '••»** 
32.  $126^ 

33.  §i|-£ 

4-  I^Iff 
5-4H 

X 

15.    $103^ 

8       4 
e"   T2T 

9-   TT7 

Page  128. 

2.   I26f 

7-  2|^ 

9«   I-20-J- 

10.  50  -J  Ibs. 
1  1.  304  A. 

I0«  *ffSr 

3.672 

9-  A 

12.  10  lots 

I  2O 

II.    -^j^y 

4-35° 
5-375 
6.  1479 

10.  4^ 

"•  i! 
12.4 

13-  42-iV4j  J- 
i4.^ 

I2O 

14-  rH^r 

7-  447f 

15.  273 

1  8.  T4j 

15.  A  bar. 

8.  1645^ 

!">•  JT^ 

400 

J9-  M 

v/         *fr  U 

1  6.  S-nnr 

9.  21630 

l6-  5-H 

Page  132. 

20.  80  days 

10.  56727^ 

17.  3^. 

i.  $43!  err. 

21.  $41^ 

1  8.  $7f 

12.135 

1  8.  7  fl- 

2. $619^0. 

22.    23!-! 

J3-  477 

ip.  6 

$19^5  er. 

23-  IIU 

Page  12  7. 

14.  266 

20.  140^  r. 

3.5o| 

24-   T% 

19.    6-^y 

15.  804 

2I<  ioff^s.r. 

4-  32f  A. 

25.  50  vests 

2°.  5-Ui 

17-  »& 

23-  f^f 

5-  II 

26.  44^  c- 

21.    if 

18.  8 

24.  ij- 

6.  3fJ 

27.  12-^  hr. 

22.    l|f 

19.   I2^| 

2  ^C     2  •Y^Vrf 

7.  $J  sold  ; 

28.  i6J 

23-  lo'oi 

2O.   I  if 

27.  T40«>A 

il  own  ; 

29.  60  bu. 

24.  12-zVff 

22.  9 

$24783 

30.  244!  yd. 

2  r       j  r  4  81 

23-  3^ 

Prtgre  131. 

8.  86^|  A. 

31.  2 

26.     15^^ 

24.  3lr 

i.  5  J  mo. 

9.  itfr 

->->       400 
3*«   T089 

*7-  20^5    25.  4^ 

2.  8|  Ibs. 

10.  f| 

33'    3384 

FRACTIONAL    RELATIONS    OF 


J'af/c  1.54. 

3-  i  A 

4-  TT 


5-   I 

6.  | 


8.  A 

10.  f  *wk. 


n. 

12. 


J-   i 


NUMBERS. 

Page  135. 

15-  $252 
1 6. 


ANSWERS. 


371 


Ex. 
I?- 

20. 

22. 
24. 

27- 
28. 

30. 
32. 


AKS. 


$132 


Ex.       ANB. 


V 

i£Jl 


33-  if 

34-  rV 

35-  W 


136, 


37-  i 

38.  I 

39-  i 

40.  I 

42-  £ 

44-  ^ 


Ex.      ANS. 


45-  A 

46.  ff  Ib. 

47.  2  ft. 

48.  A  Ib. 
49-  T* 


3-  84 

4-  9°f 


Ex.      ANB. 


6.  138 

7.  1017! 

8.  729^ 

9.  2283$ 

10.  2283* 

11.  270 

12.  $14720 
13-    33250 

16.   i/7 

17-  4 

19.  62-J 


Ex.       ANS. 


20. 


22.    288 


25.  $3  each 

27.  n|  t. 

28.  100  cts. 
8  cigars 


32.  7  times 
34-  3i 


REDUCTION     OF     DECIMALS. 

Page  142. 

Page  144. 

JPrtf/e  145. 

Page  14fi. 

8.  4-7 

3-  AV 

i.  Given 

19.  Given 

9.  21.06 

4-  TV? 

2.    -25 

20.  -3333  + 

10.  84.45 

5-  1 

3-  -4 

21.  .8333  + 

ii.  93.009 

4-  -75 

22.    .2857  + 

12.    7.045 

i 

5-  -8 

23.    .4444  + 

13.    10.00508 

a    IS 

6.  .625 

24.    .2727  + 

14.    46.0007 

°«  TO  onj 

7-  -25 

25.    .6428  + 

15.    80.000364 

9-   TTfftf 

8.  .875 

26.    .604166  + 

17.    .06;     .063; 

I0«  ?Wtr 

9.  .8 

27.    .5466  + 

.0109 

II«     I  0000 

10.  .95 

28.    75.6 

18.  .3055.00021; 

1  2.    'J  28  0~0" 

n.  .6 

29.    136.875 

.000095 

^  3-    'J'S'OiJTT 

12.    .0875 

30.    26l.68 

19.  .004;  .0108; 
.46;  .000065; 

14-    20000 
IP       S  QI7  I 

13.    .02 

14.    .000375 

31.    346.8133  + 
32.    465.0025 

.0001045 

I5-    ^30"ffO-ff 

15.    .0078125 

33.    523.00390625 

20.  69.004; 

l^'  TTT^Tnrtr 

16.  .00875 

34.    740.01375 

10.0075  ; 

I7-  tW^ff        17.  .01125 

35.    956.0078125 

160.000006 

1  8.  £ff$$£ff  ! 

ADDITION     OF     DECIMALS. 


Page  147. 

\,  2.  Given 
3.  881.6217 
4.  139.26168 

6.  118.792 
7.  892.688 
8.  2.76231 

9-  92.00537 
10.  37.417 
ii.  2.3948 

12.    23.25553 

Page  148. 

13-  575-729i°5 
14.  53.1    bn. 

^5-  75-97    lbs- 

372 


A  N  s  \v  i;  n  s . 


SUBTRACTION     OF     DECIMALS. 

Ex.          ANS. 

Ex.          ANS.           '  Ex.          ANS. 

Ex.           ANS. 

Page  14ff. 

Page  149. 

12.    .8969755 

19.    44.9955 

i.  Given              6.  7.831 

13.    .5496933 

2O.   .000098 

2.  7.831 

7.  6.60249 

14.    .876543211 

21.    $443.825 

3.  6.60249 

8-  17-3675 

15.    .01235679 

22.     139.83  A. 

4-  17.3675 

9-  77-94794 

1  6.  .099 

23.    99.063 

5.   17.94794 

10.  78.569966 

17.  .00999 

24.    70.333  m- 

ii.  2.896216 

1  8.  99.999 

io72.407m 

MULTIPLICATION     OF     DECIMALS. 


Page  15O. 

15.  43*.  25  Ibs. 

26.    .OOOO00252 

3-  -I453 

16.  $222.9375 

27.    .OOI 

4.  .000151473 

17.  469.0625  bu. 

28.   I          [ooooi 

5.  .0000016872 

1  8.  $10639.75 

29.    .OOOOOOOOOO- 

6.  21800.6 

19.  .0126 

31.    3205.05 

7.  .012041505 

20.    $686.71875 

32.    8003.56 

8.  20.08591442 

33-  2.43 

9.  318.0424 

Page  151. 

34-  5-8 

10.  721.36 

21.    .0025 

35-  5 

ii.  .00004368 

22.    .OOOOO4 

36.  $50 

12.    1.50175036 

23.    .OOOOOO49 

37.  $600 

13.    .000721236 

24.    .000000603 

38.  27.625  bu. 

14.    .O20007 

25.    .OOOOOO3 

39.  65.625  m. 

DIVISION     OF     DECIMALS. 


8.  .13+            |  16.   10 

24.  27  stoves 

Page  153. 

9-  .7^5  + 

17.  .01 

26.  4-3753 

2.    .007 

10.  .9768 

1  8.    .000002 

27.  .063845 

3-  4 

ii.  .00675 

19.  .0005 

28.  .0000253 

4.  600 

T2.    .0000576 

20.    50 

29.  .0000005 

5-  .4154  + 

13-    -015 

21.    .OOOOOO27. 

30.  $.0005 

6.  30.153  + 

14.    625.5 

22.  46  coats 

31.  $.0475 

7.  2.4142+       i  15.   10 

23.  44.409  +  r. 

ADDITION     OF     U.     S.     MONEY. 


Page  158. 

i.  Given 
2,  $48.625 

3-  $780.25 
4.  $200.09 
5-  $224.53 
6.  $761.4785 

7.  817.28 
8.  $69.1925 
9.  $13131.72 

Page  169. 

10.  $54-50 
ii.  $120.48 

12.    $76.82 

ANSWERS. 


373 


SUBTRACTION     OF     U.     S.     MONEY. 


Ex.          ANS. 

Ex.           ANS. 

Ex.          ANS. 

Ex.          Aus. 

Page  159. 

2.    $533-105 
3.    $604.625 
4.    $524.50 

5-  $585-25 

Page  16O. 

6.  $948.33 

7.   $171.9625 
8.  $411.075 
9-  $-955 

]».  $74275 
ii.  $494-945 
12.  $28.75 

MULTIPLICATION     OF     U.    S.     MONEY. 


Page  16O. 

Page  161. 

12.  $.00056 

1  8.  $45 

2.  $747-5° 

7.  $100.9125 

|  13-  $5 

19.  $80.4375 

3.  $1569.24 

8.  $14.124375 

j  14.  $.001011 

20.  $1890 

4.  $290.625 

9.  $67.8375 

15.  $6.5625 

21.  $548.625 

5.  $60.165 

10.  $310.596255 

1  6.  $58.905 

22.    $178.75 

6.  $459.25 

ii.  $4630.70025 

17.  $56.25 

23.  $14208 

DIVISION     OF 

U.     S.     MONEY. 

Page  162. 

4.  $2.75 

9.  .01 

14.    $2.965  + 

5-  $2-921 

10.    .1 

15.    925.405  + 

i.  Given 

6.   100 

II.     IOO 

16.  293.039  + 

2.  $1.964 

7.  361.455  + 

12.     IOOOO 

17.  $.0625 

3.  $.05 

8.  4000 

13.    6OOOOO 

18.  $.0547  + 

COUN1 

Page  163. 

i.  $13958.38  bal. 
2.  $159857.16  bal. 

r  I  N  G  -  R  O  O  M 
Page  165. 

2.    $206.83 
3.    $4367.125 

EXERCISE 

4.   $3183.07$ 
Page  166. 

5.  $71.15 

:s.  . 

6.  $5201.70 
7.  $717.77 
8-  $395-37 

ANALYSIS. 


Page  167.    !i3.  $114.06$ 

20.  6132  A. 

10.  $31! 

4-  $789^25 

14.  $47.82  ami 

21.    $478.125  1. 

ii.  $291! 

5.  $4312.50 

15.  $3007750 

2.  $66J 

12.  $35 

6.  $43l 

16.  946.  i+lb. 

3-  $65 

13-  $27! 

7.  $892* 

17.  7897  Ib.  s'd; 

4-  $22 

14.  $107$ 

8.  $0.33^ 

149  Ib.  av. 

5.  $409$ 

15.  $115$ 

9-  $5-264 

1  8.  36.45  bu. 

6.    $12$ 

10.  $31.50 

7.    $!50 

Page  169 

ii.   102  bibles 

Page  16  S. 

8.  $208$ 

17.    $210 

j2.   148^  vests  i  19.  $247 

9-  $58if 

18,  $547j 

374 


ANSWERS. 


Ex.          ANS. 

Ex.         ANS. 

Ex.          ANS. 

Ex.          ANS. 

I9.    $576 
20.  $168 

21.    $240 

23.  13.66  yds. 
24.  1125  Ibs. 

25.  1800  coc's 
26.  2690  pine's 

Page  17  O. 

2.    $78.177 

3.   $240.625 
4.  $789.625 
5.    $1169.83-! 
6.  $1203.93! 

7.  $40.4875 

8.  $22.31} 

9.    $100.2472 
10.  $4.51 

II.  814.18-} 

12.    $15.975 

REDUCTIO  N. 

fage  WO. 

Page  192. 

55-  *'2-75 

3.  614  far. 

30.  495782  in. 

56.  790  gi. 

4.  39618  far. 

31.  126720  in. 

57.  603  qts. 

6.  £8,  IDS.  7<1. 

32.  $1480 

58.  5713  qts. 

7.  128.  gel.  2  far. 

33.  $1556.10 

59.  34616  gi. 

8.  £41,  53.  2cl.  2  far. 

34.  456  eighths 

60.  10834  gal.  i  pt 

9.   1  66  1  yds. 

35.  2608  sixteenths 

6  1.  1  68  bottles 

(o.  £225 

36.  i44J  yds. 

62.  $126.00 

ii.  4440  pwts. 

37.  123!  yds. 

63.  2612530  sec. 

38.  22  vests 

64.  176010  min. 

Page  191. 

39.  $16.25  profit 

65-  3I556929-7  sec. 

12.    7865  pWtS. 

40.  1  60  r.  256  sq.  ft. 

66.  gw.  6d.2oh.i5m. 

13.  80  Ibs.  7  oz. 

41.  5  A.  405.  r.  2os.y. 

67.  639  y.  225  d.  5  h. 

14.  7  Ibs.  i  oz.  iSpwt. 

42.  6976414}  sq.  ft. 

68.  19  y.  164  d.  6  h. 

1  8  grs. 

43.  10240000  sq.  r. 

20  min. 

15.  1491  rings 

44.  10  A.  1  08  sq.  r. 

69.  $212.625 

1  6.  $27 

45.  $22875  profit 

70.  13°  29'  21" 

17.  2653  oz. 

46.  3983040  en.  in. 

71.  365.7°  17' 

1  8.  6721944  oz. 

47.  32000  cu.  ft. 

72.  855631" 

19.    2  t.  725  11). 

48.  10  c.  ft.  985  c.  in. 

73.  1296000" 

2O.    II  t.  1141  Ib.  I2OZ 

49.  64  C.  86  cu.  ft. 

74.  163^  dozen 

21.    $700.55 

50.  5259  qts. 

75.  1  500  eggs 

22.  $280  profit 

51.  24051  pts. 

76.  694!  gross 

23.  24048  drams 

77.  9360  pens 

24.  7lb.6oz.  idr.  2sc. 

Page  19.'}. 

78.  67  Ibs. 

26.  1928495  ft. 

52.  6641  bu. 

79.  1800  sheets 

28.  7  in.  79  r.  i^  ft. 

53.  254  bu.  2  p..3  q. 

80.  4i6f  quires 

29.  443440  rods 

54.  $412.08 

APPLICATIONS    OF    WEIGHTS     AND     MEASURES. 

Page  194.         Page  19fi.        7.  2  sq.  ft          10.   192  bulbs 

i.  Given               4-72  rods          8.  1200  bu.       n.  3630  gr.  v. 

2.  i\  A.               5.  $46500  pr.          $300                   $190.575  & 

3.  165  ft,             6.  4116  s.  ft      9.  Given           12.  z\  A. 

ANSWERS. 


375 


Ex.          ANS. 

Ex.           ANS. 

Ex.          ANS. 

Ex.          ANS. 

Page  196. 

23.  147  b.  ft. 
24.  $278.10 

35.    $112 
36.    $337.68 

47-  i795ff  g- 

14.  $324 

37.  86751  br. 

Page  2O1. 

15.  $8.88 
16.  $402.27! 
17.  $22.50 
18.  $45*139 

Page  198. 

28.  $14.85 
29.  $23.40 

38.  $3600 

Page  2OO. 

40.  56  yds. 

48.  56^  in. 

49-  5  7-85  7  +b. 
50.  42  1  Ac-  ft 

51.  9-9551  in- 

*2O«    5P4.R 

41.  44^  yds. 

52.  3750  Ibs. 

Page  197. 

Page  199. 

42.  12^  yds. 
43.  320  sods 

C  ^,      <tp7T^TTT 
•J  -J         ^  /    L  £  o 

56.  9y5^  Ibs. 

21.  i8fb.fi; 

32.  2-J-^f  cords 

44.  48  yds. 

57.  87!  rings 

$1.40  val. 

33.  i6|  tons 

45.  640  tiles 

58.  $72-f| 

22.    12^  b.  ft. 

34.  $133.20 

46.  1  3^  rolls 

59.  $805^1 

DENOMINATE     FRACTIONS. 


Page  202. 

Page  2O4. 

30.  |^|  wk. 

41.  17.28  grs. 

2.  $  qt. 

1  6.  133.  4d. 

31.  |  C. 

42.  £5,  i2S.  6d. 

17.  3  pk.  6  qt. 

32-  ff 

0.4032  far. 

3-  "JT  u« 

1  8.  1250  Ibs. 

4.  -fe  br. 

19.  10  oz.  10  p. 

Page  2O5. 

Page  2O6. 

5-  3f  °z- 

20.  3  fur.  13  r.  i 

34.  2S.  6d.o.4  + 

44.  .5423  +  lb. 

6.  ^  s.  in. 

yd.  2  ft.  6  in. 

far. 

45.  .005  ton 

J 

21.  146  s.r.  2os.y. 

35.  10  oz.  iSp. 

46.  .45539  +  m. 

i  s.  ft.  72  s.in. 

22.56  grs. 

47.  .04  lb. 

Page  2O3. 

22.    112  CU.  ft 

36.  2  furlongs 

48.  .5  bl. 

g.     Sy^ 

24.  ££  gal. 

32  rods. 

49.  .409+  r. 

Q       £        T 

25.  i-i|?  lb. 

37.  2  qt.  i  pt. 

50.  .25 

10.  Tt^oz. 

26.  fl  bu. 

38.  39^36" 
39.  i  d.  i6b.29 

51.  .48  bl. 
52.  1.44  r. 

ii-  ritfgal- 

27.  ^ff  ton 

•  m.  1  6.8  s. 

53.  .166+  wk. 

12.    -f-SV  m- 

28.  ff£  sq.  yd. 

40.  505  lb.  1.92 

54.  .28182  + 

'3-trfwt. 

29-  tWl 

oz. 

55.  .41666  +  0. 

METRIC     NOTATION     AND     NUMERATION. 

Page  212. 


5370.9845  dekameters; 
537.09845  hektometers ; 
53.709845  kilometers; 
537098.45  decimeters. 


2.  450.5108  dekagrams; 
450510.8  centigrams; 
45.05108  hektograms ; 
4.505108  kilograms. 


37G 


ANSWERS. 


REDUCTION     OF     METRIC     WEIGHTS    AND     MEASURES. 


Ex.                  Ass. 

Ex.             Aus. 

Ex.                ANS. 

rage  213. 

i.  Given 
2.  437500  sq.  m. 
3.  867000  grams 
4.  26442  liters 

5.  256100  sq.  m. 
6.  8652000011.  dm. 
7.  4256250  grams 
8.  Given. 
9.  65.2254  hektars. 

10.  0.087  kilos, 
u.  1.48235  kg. 
12.  39.2675  kl. 

APPLICATIONS     OF    METRIC     WEIGHTS    AND     MEASURES 


Page  215. 

8.  Given 

15.  72.492+  kl. 

3.  39.14631  m. 

9.  148.87775  A. 

16.  143.223+  kg. 

4.  1  9.8  134  gals. 

10.  4237.92  cu.  ft. 

17.  6000.06+  s.  m. 

5.  15.89  bu. 

12.  58.2934-  m. 

18.  16.378+  beet 

6.  4.2324  oz. 

13-  6236.959+  kg. 

19.  410.748+  c.  m. 

7.  303.68365  Ibs. 

14.  236.5  85  +  liters. 

20.  27958.715  +c.m 

COMPOUND    ADDITION. 

Page  217. 

9.  1  96  bu.  2  pk.  7  q. 

14.  64  A.  7  s.r.  i  o-J  s.y 

3.  £11,  xos.  od.  2  f. 

10.  6  C.  80  cu.  ft. 

17.  15  cwt.  50  Ib.  4oz 

4.  z6T.3Cwt.83lb. 

ii.  98  bu.  3  pk.  2  q. 

1  8.  2  pk.  4  qt. 

3  oz. 

19.  goz.  ipwt.  logr. 

5.  45  bu.  o  pk.  2  qt. 

Page  218. 

20.  3  s.  6  d.  3.2  far. 

6.  74$  yds. 

12.  Given 

21.  2  A.  52  sq  r. 

7.  1093  Ib.  5  oz. 

13.  23  wk.  i  d.  17  h. 

22.  2  C.  8  1  en-  ft» 

8.  55  gal.  2  qt. 

58  m. 

1209!  cu.  in 

COMPOUND    SUBTRACTION. 

Page  219. 

10.  3  m.  5  fur.  38  r. 

4.  150  days 

5  yd.  o  ft.  i  in. 

5.  224  days 

2.  i  fur.  39  r.  i  yd.  2^  ft. 

2."  9  s.  9  d. 

6.  10  1  days 

3.  7  Ib.  4  oz.  1  7  p.  9  grs. 
4.  7  T.  8  cwt.  26  Ib. 

'3.  2pk.sqti.2p. 

Page  223. 

5.  8  gal.  i  qt.  i  pt.  2  gi. 

4-  4^  pt- 
5.  i  jib.  2  oz.  2  p. 

10.  86°  19'  24" 

o            /        // 

6.   ii  4.  4  sq.  r. 

11.  7    24'  7 

OO          ' 

Page  22O. 

7.  31.25  Ibs. 

12.    10     2 

13.  5°  6'  46" 

6.  159  A.  12  sq.  r. 

Page  221-2. 

14.  15°  4'  16" 

222^  sq.  ft. 
7.  46  cu.  ft.  1689  cu.  in. 

3.  67  y.  gm.  22  d. 

15-  19°  35' 
16.  56°  u' 

8.  145  A.  21  sq.  r, 

2.  113  davs 

17.  70°  29' 

9.  46$  yd.                      1  3.  74  days 

18.  10°  19'  38" 

ANSWERS. 


377 


COMPOUND     MULTIPLICATION. 


Ex.                AN-. 

Ex.            ANS. 

Ex.               Ass. 

Page  224. 

2.  98T.i7Cwt  28  Ib. 
3.  £151,  153.  9fd. 
4-  33oz.ispw.iog. 

Page  225. 

7.  331  gal.  2  qt 

8.  22  C.  91  cu.  ft 
9.  £23,  153.  3^(1. 
10.  562m.  4  fn.  24  r. 
ii.  1937  bu.  i  pk. 
12.  22  C.  57  cu.  ft. 
13.  6  1  T.  844  Ibs. 
14.  1307  r.  8  qr.  8  s. 

15-  161°  37'  30" 
16.  431  h.  15  m. 
!?•  J33  sq.  r.  21  sq. 
yd.  £  sq.  ft 
18.  73  T.  1492  Ibs. 
19.  1571  bu.  2  p.  4  q. 
20.  1  946  gal  3  q.  i  p. 

COMPOUND     DIVISION. 


Page  227. 

3.  4  fur.  8  r.  2  yd.  2^  ft 
4-  6  gal.  3  qt  o  pt  3!  gi. 

5.  25  bu.  o  pk.  i  qt  I  pt. 

6.  15  A.  1 06  sq.  r.  5  sq.  yd. 
sq.  ft 


9- 
10. 


26f  spoons 
8800  rails 
1 049^  times 
12  books 


11.  2  A.  64  sq.  r. 

12.  6  bu.  i  pk.  i  qt. 


COMPARISON     OF    TIME    AND     LONGITUDE. 


Page  228. 

Given. 

43  m.  32.13+  s. 

1 1  o'c.  1 2  m.  4  s. 

1 2  o'c.  3  701. 8.43. 
49  m.  20  s. 


6.  i  o'c.  30  m.  E. ; 
10  o'c.  30  m.  W. 

7.  54  m.  19.8  sec. 

8.  13  m.  44.86+3. 

Page  229. 

9.  Given 

PERCENTAGE. 


10.  6°  9' 

11.  6°  45' 5" 

12.  45°  24'  45" 

13.  56  m.  49+  sec, 

14.  12  o'clk.  34  m. 
47  sec. 


13 
14 
'5 
Page  234. 

3.  $24.21 

4.  10.8  bu. 

5.  22.6  yds. 

6.  8.64  oxen 

7.  8.25  yds. 

8.  $11.584 


Page  232. 

•5°;  -2s;  -75;  -2°;  4°;  -60;  .80 
.10;  .70;  .90;  .05;  .35;  .12;  .06 

.66f;  .i6|;  .125;  .625;  .875;  .58$;  .91$ 


9.  100.8  Ibs. 
10.  63.14  T. 
IT.  52.5  men 
12*.  145.656  m. 

13.  $857.785 

14.  £.106 
15-  1-25  1- 


1 6.  1.38  k. 

17.  840  Ibs. 

18.  81000 

19.  78.75  bu. 

20.  .35  Ibs. 

21.  9.765  gals. 

22.  840  men 


24.  $215 

25.  157.2  Ibs. 

26.  32.25  m. 

27.  116  1. 

28.  580  sheep 

29.  156  bu. 

30.  $925 


378 


ANSWERS. 


Ex.          ANS. 

Ex.          ANS. 

Ex.          ANS. 

Ex.         ANS. 

.ji.  576  cu.  ft. 

14.   $10200 

Page  238. 

19.  3000  m. 

32.  £2,  is. 

15-  $9750 

4.  250  bu. 

20.  360000  p. 

5.  $362.50 

Page  236. 

Page  237. 

6.  1  80  tons 

Page  24O. 

3.  1150  s.  C.; 

2-r  -7  I  £7 
*     o^r/^ 

7-  £45° 

5.  809 

880  s.  D. 

3-  25$ 

8.  600 

6.   1820 

4.  $2300.91 

4.  33$$ 

9.  360  fr. 

7.  Given 

5.  $2850 

5.  15% 

10.  8000 

8.  f 

6.  504  bales 

6.  t)\% 

II.    $200 

9-  A  B's 

7.  2018.75  gal. 

7-  S's/o 

12.  $i6666f 

10.  5175  m. 

8.  1312  m. 

8.  10% 

13.  $17600 

ii.  $3648 

9.  $470-40 

9.  80^ 

14.  54.4  yds. 

12.    6875  P- 

10.  129  t. 

10.  10% 

15-  $50 

13-    $3350 

ii.  $1410.50 

II.  23-^% 

16.  100 

14.  228  sheep 

12.    $228 

12.    42^ 

17.  $5000 

1  5.  400  pupils 

13.    $4823.33  J 

X3-  3°fj/£ 

18.  $1920 

1  6.  5800  m. 

COMMISSION     AND     BROKERAGE. 


Page  242. 

3.  $31.1431$; 
4.  $366.225  com.; 
$10902.225  paid 

Page  243. 

5.  $4310.145 
6.  $238.66-}; 
8.  \% 
9-  5% 
10.  \% 

Page  244. 

12.    $9000 

13.  $4212  sales; 
$4001.40  net  p. 
14.  $1500  col.; 
$1432.50  paid 
15.  $9000  sale  ; 
$8865  rec'd 
17.  $5100 
1  8.  $6834.87  2  sales 
$170.872  com. 

Page  245,  6. 

19.  $15488.89  + 

20.  $26635.294  + 

22.  $3010.75 

23.  $2419.23  + 

24.  #49261.083  +in.; 
$738.916+  com. 

25.  $13687.50  g.  a.; 
$360.75  ch.; 
$13326.75  net  p. 

26.  $12127.8?  net  p. 


PROFIT    AND     LOSS. 


Page  247-9. 

ii.  $453.60 

19.  $12810 

3.  $22.20 

12.    $2942.50 

21.    28f% 

4.    $22 

13.    $9520 

22.    11^% 

5.    $.0875 

14.  1  1£  cts. 

23.  6o<£ 

6.  $16.65 

15.  $i  a  pair 

'25.   100% 

7.  $1.40 

1  6.  $2.77!  each 

26.    50^ 

10.  $2925  A.; 

17.  $4.90  a  yd. 

27.  66$% 

$2075  B. 

1  8.  $4331.25 

28.  40^ 

A  X  S  W  E  E  S . 


379 


Ex.              ANS. 

Ex.              ANS. 

Ex.             ANS. 

Page  25O. 

Page  252. 

14.  $7  cost; 

29.  50% 

51.  $.736  marked  p. 

$8.05  sell'g  pr. 

30.  Given 

52.  $34.78+  m.  pr. 

15.  $2  cost; 

31-  s°% 

53.  $4.80  m.  pr. 

$1.75  sell'g  pr. 

32.  5°$ 

54.  $i.i6f  m.  pr. 

33-  33i# 

55.  $75  each 

Page  254. 

34.  $12.60  pr.; 

56.  $5.75  per  yd. 

1  6.  $5200  sales; 

7H#    ' 

$4940  net 

36.  $9750 

Page  253. 

17.  $24000  sales 

i.  $2.691  prof. 

18,  $1.25  cost 

Page  251. 

2.  $6356.72^ 

19.  $5250  cost; 

39.  $i  cost  per  Ib. 

3.  $3487.575  sh.; 

$1750  gain 

40.  822500  cost; 

$850.84  ging.  ; 

20.  .7894^  per  gal.; 

$30000  sell'g  pr. 

$4338.415  both 

.  1  48  1  gain  per  g.; 

41.  $17500  spend; 

4.  $22  loss 

$49.73  wh.  cost; 

$2  1  ooo  am  t.  sales 

5.  $77812.50  cost 

$9.33  wh.  gain 

42.  $12000  A's  inv.  ; 

6.  $897.45  net  pr. 

21.  $10  a  barrel 

$9375  B'sinv. 

7-  38f# 

22.  $51311!  cost; 

43.  8250000  cost; 

8-  56^  prof.; 

$6413!  gain 

$275000  amt.  s. 

$3240  prof. 

23.  $157  98.963  sale; 

44.  $.02  cost; 

9.  i6f%  loss 

$55  2-963  com. 

$025  selling  pr. 

10.  .052  or  ST%-#C.; 

24.  3*&% 

45.  Given 

$1766.835  net  p. 

25-  5Tn&#  loss 

46.  $10061.  71} 

ii.  25$  loss 

26.  $60  m.  p. 

47-  $13319-672 

12.  125$ 

27.  (>\%  loss 

48.  £22  cost 

13.  $9300  cost; 

28.  15  cts. 

49.  3  cents 

$11  1  60  sell'g  pr. 

29.  lo-j^ib  loss 

INTEREST. 

Page  258. 

14.    $680.28 

27.  $99.96         |    6.  $73.96 

3-  $2.54 

15.    $81.44  + 

28.  $24.00             7.  $64.18 

4.  $1.676 

16.  $2875.792 

29.  $163.842  ;  I    8.  $26.49 

5.  $1.224 

17.  $362.315 

$2740.652   !    9.  $1.65 

6.  $4-523 

18.  $2759.962 

10.  $29.933 

7-  $43-356 

19.  $5032.083 

Page  259. 

1  1  .  ^60.04 

8.  $110.77 

-20.    $122.40  £ 

2.    $24.44                I2«    $30.52 

9-  $33-42 

21.    $226.69-^ 

3-  $6.252 

13.  #2450.80 

10.  $23.364 

22.    $45.053 

14.  $20819.80 

11.  $25.577 

23.    $216.489 

Page  2(iO. 

12.    $94.35 

24.    $493.20 

4.   §4.82 

Page  261. 

13.    $17,91           ^25.    $3923.47 

5.  $6.75 

3.  $2.8435 

8U 


ANSWERS. 


Ex.          ANS. 

Ex.          ANS. 

Ex.          AKS. 

Ex.          ANS. 

4.  $2.192 

4-  <>% 

7.  i4f  jrs. 

2.   $235.85 

5-  $11-413 

5-  8# 

8..i6f  yrs. 

3-  $327-36 

6.  $13.89 

6-  9&% 

9.  i8|  yr-s. 

4.   $8928.57 

7.  $1285.33          7-  *\% 

5.   $892.86 

8.  $13.876            8.  s%\  ™% 

2.  $i666f 

6.  $5582.142 

9-  $4363-044 

9-  7$ 

3.  $3000 

10.  $H4.i6|- 

10.  6$ 

Page  268,  .<>. 

ii.  $185.18 

Page  264. 

i.  Given 

12.    $81.358 

Page  263. 

4-  $2333$ 

2.    81519.71 

3-  5f  .Y->  or  5 

5.  $2500 

3-  *5  38-63 

Page  262. 

y.8m.i7d. 

6.  $40000 

5.  $270.19 

2.    1% 

4.  9  m.  2  d. 

7.  $10000 

6.  8388.23 

3-  1% 

5.  £y.,or3m. 

8.  $25000 

7.  $516,32 

COMPOUND     INTEREST. 


Page  273,  4. 

5.  8200.63 

8.  $551.58 

13.  $21825.26 

i.  Given 

6.  82165.713 

9.  $1377-41 

14.  $29849.56 

2.    $112.52 

10.  $1543-65 

15-  $37/04.95 

3.    $161.63 

Page  275. 

ii.  $8104.25 

1  6.  $3298.77 

4.    $164.61 

7.  Given 

12.    $32564.58 

- 

DISCOUNT. 

Page  277. 

7.  Given 

I3.    $528.33 

6.  8'  17.*$^  i 

i.  Given 

8.  886.57 

14.  846,73  for. 

7-  $735-7° 

2.    $283.47  + 

9.  8101.892 

X 

8.  Given 

3.    $462.96 

10.  891.35  dis; 

Page  279. 

9.  8768.44  -^ 

4-  $1213-59 

82  283.65  p. 

3.  $718.685 

10.  $15787*4. 

5.  $2336.45 

ii.  $3775.264 

4.  8989.50 

ii.  $2226.23 

6.  84464.28 

12.  $20.377 

5.  $1721.875 

12.  $7503.83 

STOCKS    AND    BONDS. 

Page  282. 

8.  $12880 

12.   $10852.37$ 

Page  284. 

2.    $243 

13.    $19293.75 

20.  40  bonds. 

3-  $525 

Page  2S3. 

14.  $672  cur. 

21.  $12000 

9.  $10497.50 

1  6.  15% 

23.  $44166$ 

5.  8400 

10.  $6363.61 

17.  6\% 

24.  835625 

6.  $476 

ii.  $6336.11 

1  8.  6%% 

25.  $56000 

ANSWERS. 


381 


EXCHANGE. 


Ex.          Asa 

Ex.          ANS. 

Ex.          ANS. 

Ex.           ANS. 

Page  287. 

2.    $2049 
3.    $3488/80 
4.    $4130.647 
5.    $203 

8.  $1219.51 

Page  288. 
o.  $1491.053 

10.  $2617.801 
II.  $3751.95 
12.    $3750 

Page  289. 

1,2.  Given. 
3.  $1858.80! 
4.  $4866.50 

Page  29  O. 

6.  $56125.12  + 
7.  Given. 
8.  £5  1  1,  15  s. 
4  d.  3.2  far. 
9.  £774,  7  s. 
10  d.  i.  28  far. 
10.  £1026,  8s. 
7.2  d. 

1  1.  .£1542,1  is. 

6d. 

Page  291. 

12.  Given 
13.  $675.68 
14.  Given 
15.  13050.00  f. 
1  6.   16474.50  f. 

INSURANCE. 


Page  293. 

i.  Given 
2.  $6.65 

3-  $38-40 
4.  $84.375 

5.  $10.70 
6.  $25.00 
7.  $197.60 
8.  $875 
9.  $1569.625 

Page  294. 

10.  Given 
ii.  $15747.423 

12.  $27806.122  + 

*3-  $37105.263  + 

i.  Given 

2.  $125.00 

3-  $2437.50. 
4.  $5000 
5.  $45000  gr. 

Page  297. 

3.  $1003.75,  C'stax 
$1250,      D's  "' 
$1375,      E's  " 
$925,        Fs  " 


TAXES. 

4.  2%  rate; 
$159.75,  A's  tax 

5.  z%  rate; 
$450,  G's  tax 

6.  $309.75  H's  tax 


7.  $%  rate; 
$170,  A's  tax 

9- 
10. 

ii.  $11052.63  + 


Page  299. 

2.    $4062.50 
3.    $1063.314 

DUTIES. 
4.   $1999.20 
6.  $945  v 
7.  $2296.80 

Page  3OO. 

2.    $Il8.25 
3.    $425.75 

EQUATION     OF     PAYMENTS. 


Page  :t(>2. 

2.  4  m.  from  J.  20,  or  Oct.  20 

3.  8  ra.  18  d. 

4.  Mch.  10+17  d.=Mch.  27 

5.  June  i,  or  in  3  in. 

AVERAGING 
Page  3OO. 

1-3.  Given 
4.  7  m.  extension ; 
Bal.,  $1650 


Page  803. 

6.  Given 

8.  July  15+51  d.— Sept.  4 

9.  $1483.25  ami; 

42.9  d.  av.  time ;  due  Nov.  1 7 

ACCOUNTS. 

5.  $300  btil.  debts; 
22320  bal.  prod.; 
74  d.  av.  time 
Due  May  23. 


38S 


A  N  6  W  E  B  S . 


SIMPLE     PROPORTION. 


Ex.          AN*. 

Ex.           ANS. 

Ex.          ANS. 

Ex.          Ana. 

Page  312. 

14.   8133 

26.   85223^ 

38.  25  min. 

•?.  8160 

15.  2  240  times 

27.   812.50 

39.  6600  rev. 

O 

4-  8165 

1  6.  5  years 
17.  8  hours 

28.  14.4  in. 
29.  26f  yds. 

40.  840 
41.  2\  hrs. 

6    &6 

30.  97920  t. 

42.  37?y  hrs. 

7.  £19,  7s.6|d. 

Page  314. 

18.  1  100  men 

31.   180° 
32.  80  ft. 

43.  45  days 
44.  1000  rods 

19.  240000  Ib. 

33.  64^  orang. 

45.  300  hrs. 

-P«<jre  3  13. 

20.  27  horses 

34.  3000  mi. 

46.  8191.78 

9-  i35  A. 

21.  8612 

35.  33  men 

47.  8m.  518. 

10.  8937^ 

22.    £l£ 

36.  30  ft. 

before  9 

ii.  270  miles 

23.  85.06^ 

48.  3^  cts.  loss 

12.  8288 

24.  8880.40 

Page  315. 

49.   1  50  m.  less; 

13.  8822^ 

25.  8490-5  2i 

37.  $1920 

1  86  m.  gr. 

COMPOUND     PROPORTION. 

Page  317. 

Page  3  IS. 

9-  $75° 

13.  750  Ibs. 

3.  2$  days 

6.  Si26.66f 

10.  10285^  yd. 

14.  862.50 

4.  9  horses 

7.  $100 

ii.  37-Jdays 

15.  240  sofas 

5.  331  weeks 

8.  360  miles. 

12.  &I454A 

1  6.  864  tiles 

PARTITIVE 
Page  319. 

40  s. ;  60  s. ;  100  s. 


PROPORTION. 


4.  60  bu.  oats;  Sob.  p.;  nob.  c. 
5-  7i;  !o6|;  142;  177!  A. 


2.  $222$,  A's  loss; 
$277^,  B's  " 

3.  8340,  A's  share  ; 
8510,  B's   " 


PARTNERSHIP  .—Page  321-3. 

896B's;$i6oC's    n.  81500,  H'd.; 
8.  8i42.8q4,A'ssh.          82250,  Cont.; 
83000,  Am. 

12.  8263.38^]  |,A's; 


4.  $5  06  7  1|,  A's  g'n; 


-H,  C's  « 

5.  $3400;  $5100; 
and  86800 

6.  864,  A's  share  ; 


Page  324. 

2.  81687.50,  A's  sh.; 


$266.66f,  B's  " 

8  190.47!!,  C's" 


9.  824.48!!,  one  ; 
$25.51^,  other 
10.  8424^,  A's  sh.; 
8424^,  B's  u 
C's  « 

,  D'S  " 


13.  8402 1 -f ||,  A's; 
$3846!!!,  B's; 


BANKRUPTCY. 

81205.35^,  B's  sb.; 
8857.14!,  D's    « 


3.  82060,  D  rec'd ; 
^12100  net  proc. 


AN  S  WEES. 


383 


ALLIGATION. 


Ex.          ANS. 

Ex.           ANS. 

Ex.          ANS. 

Ex.           ANS. 

Page  323. 

2.    I5|  Cts. 

9.  2  p.  at  15  ; 
i  p.  at  1  8  ; 

Page  328. 

1  8.  54-j^lb.  280; 
iS-^j-  Ib.  30  ; 

3-  95A  cts. 

2  p.  at  2  1  ; 

14.  3-J  Ibs.  ea. 

54iT  -Ib.  3"  5 

4-93?  cts- 
5.  2o££  carats 
6.  $15.68 

Page  327. 

8    T.  <5  bu  i  st  * 

5  p.  at  22 
10.  9lb.at32  c; 
61b.  at  40; 
6  Ib.  at  45 
ii.  81b.at2oc; 

15.  5oq.at4C.; 
300  q.  at  6 

Page  329. 

17.  2olb.  at6c.; 

72-frlb.  42 
For  other  ans. 
see  Key. 

53i  g-  45  5 

40  bu.  2d  ; 
40  bu.  3d 

5  Ib.  at  35  ; 
1  2  Ib.  at  40 

20  Ib.  at  8  ; 
60  Ib.  at  12 

140  g.  60 

INVOLUTION  .—Page  331. 

8.  25;  36;  49;  64;  81 ;  100 ;  400;  900;  i6oa;  2500;  3600, 

4900;  6400;  8100. 

9.  .25;  .36;  .49;  .64;  .81;  .0001;  .0004;  .0009;  .0016;  .0025; 

.0036;  .0049;  .0064;  .0081. 


10.  125 

11.  64 

12.  2299968 


13.  1024 

14.  4096 
IS-    IS625 


16.  8.365427 

17.  64.014401080027 

1 8.  64024003.000125 


21. 


EXTRACTION     OF    THE     SQUARE     ROOT. 


Page  338.  12.  .8514  + 

20.  3.16  + 

28.  98.7654 

37.  4-1683 

5-  427 

13-  -355  + 

21.  3.316  + 

38-  i£ 

6.719        (14.1.635  + 

22.3.46  + 

Page  339 

^o.  24 

7.772 
8.  1.892  + 
9.  .64 

15.69.47  + 
16.  21.275  + 
17.  2.236  + 

23.  4.42 

24-  57-3 
25.  9.36  + 

33-  -745  + 
34.  .866  + 

40.  .8545  + 
41.  i.or8  + 

10.  .347 

18.  2.64  + 

26.  1  1  ii  i.  ii  + 

35.  2.529  + 

42.  -V- 

ii.  .41          ;  19.  2.828  + 

27.  7856.4 

36.  3.63  + 

43-  ¥ 

Page  339-41. 

2.  420  rods. 
3-  559-28  r. 

4.  119  trees 

5.  238  men 

7.  1 08  yds. 

8.  30  rods 


APPLICATIONS. 


9.  438.294-  m. 
10.  103.61  +  ft. 
ii.  56.56+  ft. 
12.  75.81+  ft 

1  6.  Given 
17.  12 

18.  54 
19.  63.49  + 

Page  342. 

20.  1.75 

15.  3^  min.                |    22.  •£& 

AK8TTEBS. 


EXTRACTION     OF     THE     CUBE     ROOT. 


Ex.         AKS. 

Ex.          ANS. 

Ex.          ANS. 

Ex.          ANS. 

Page  X4S. 

1-2.  Given 

3-  85 

4.  4-38  + 
5-  72 
6.  1.44  + 

7.    2.76 

8.  2.57  + 
9.  8904 
10.  .632  + 
ii.  85.6+  yds. 
12.  15.3+  ft. 

13.  129.07  -fin 
14.  Given. 
16.  .7464- 

!*..» 

18.  A 
19.  4-3  + 

Page  34ft. 

i.  Given 
2.  52  ft. 
4.  1  6  ft. 
5.  462.96  +  c.f. 
6.  3.072  tons 

ARITHMETICAL    PROGRESSION  .—Page  351. 

I-   IS 
3-  8 

5.  $360 
7.  ii  children 

Page  3~»2. 

9-  3  FS. 

10.  Given 
ii.  78  strokes 

GEOMETRICAL 


Page  353. 

1.  96 

2.  $10.24 


§2007.3383664; 

£3001.460703698 

242 


PROGRESSION. 

Page  354. 

5.  1456 

6.  $4095 


MENSURATION  .—Page  35<i-9. 


i.  Given 

9.  204.20335  r. 

15.  3817.03185  cu.  ft 

2.  3600  s.  y. 

10.  47.746-mm  ft- 

I  6.    522f  CU.  ft. 

3.  1  44008.1'. 

TT         •?  T    5\  ^  T       4-8                  i« 
A             O            O           14159        * 

1  8.  756  sq.  ft. 

4.  80  rods 

12.    4417.8609375  S.  ft 

20.  14684558.207965.  m. 

7.  $90.90-^ 

13.  3183.1  s.  r. 

22.  260651609333^00.  m. 

MISCELLANEOUS    EXAMPLES  .—Page  3GO-2. 

I.  $1083! 

$45.93^  pr. 

21.  150  rods 

36.  6  1  ft 

2.    $120  COSt 

5 

12.    12  O'cl.  31 

22.    £ 

!ft. 

37.1226196.489 

$45  loss 

m.  2o-fT  s. 

23.  $6038.72 

38.  44% 

3.  692  less: 

13.  $120  cost; 

24.  45°  24'  45" 

39.  $2607.291 

1434  great 

$30  profit 

25.  $8125 

40.  $5400.  ist.  ; 

4.    22  ft. 

14.  35  m.  8  s. 

26.  £3f| 

$8100,  2d.  ; 

5.  240  days 

past  7  A.M. 

27.  $16836.734 

$13500,  3d 

6.  143^  miles 

15.  $24 

28.  14520  per. 

41.  82574.50 

7.  $20,  A's; 

1  6.  3  farmer's; 

29.  $19878.92 

42.  $50,  A; 

$40,  B's  ; 

24  neigh  b's 

30.  $680.625 

$75,  B; 

$140,  C's 

17.  540  sheets 

31.  77*  ft. 

$105,  C 

8-  79-591  ku 

. 

1  8.  $40,  A's; 

32.  $12500 

43.  Si  20  each 

9.  5.26T2j  C. 

$60,  B's 

33-  * 

1.59 

$130,  B 

10.  $3 

19.  1980  pick. 

34.  $1446.625 

44.  5  hrs. 

ii.  $164.06$ 

c. 

20.  15^             i  35.  48  ft, 

45.  $265720 

o 


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